LIBRARY 


THE  UNIVERSITY 
OF  CALIFORNIA 

SANTA  BARBARA 

PRESENTED  BY 
Glen   G.    Mosher 


GLEN     G.     MOSHER 


PRACTICAL 
STRUCTURAL  DESIGN 


A  TEXT  AND  REFERENCE  WORK  FOR  EN- 
GINEERS, ARCHITECTS,  BUILDERS,  DRAFTSMEN 
AND  TECHNICAL  SCHOOLS;  ESPECIALLY  ADAPTED 
TO  THE  NEEDS  OF  SELF-TUTORED  MEN 


BY 

ERNEST  McCULLOUGH,   C.E. 

MEMBER   AMERICAN  SOCIETY   CIVIL  ENGINEERS,'    CONSULTING 

ENGINEER;  LICENSED  STRUCTURAL  ENGINEER  AND 
LICENSED  ARCHITECT,  STATE  OF  ILLINOIS 

AUTHOR   OF   "ENGINEERING  AS   A  VOCATION,"    "PRACTICAL 

SURVEYING,"  "ENGINEERING  WORK  IN 

TOWNS  AND  CITIES,"  ETC. 


U.  P.  C.   BOOK  COMPANY,   INC. 

243-249  WEST  39-TH  STREET 

NEW  YORK 

1918 


COPYRIGHT,    1917,    BY 
U.   P.   C.   BOOK   COMPANY,   INC. 


PREFACE 

COMMENCING  in  the  year  1914  a  series  of  articles  by  the 
author  of  this  book  appeared  in  the  pages  of  Building 
Age,  with  the  title  "  Design  of  Beams,  Girders  and  Trusses." 
The  articles  were  completed  early  in  1916,  and  in  reply  to  an  in- 
sistent demand  they  were  prepared  for  appearance  in  book  form 
with  practically  an  equal  amount  of  new  material.  The  present 
book  is  the  result. 

Before  writing  the  articles  the  subject  matter  had  been  tried 
out  on  a  number  of  classes  of  students,  some  of  them  in  evening 
schools  and  some  in  private  classes  organized  for  the  purpose  of 
preparing  the  students  to  pass  state  examinations  to  obtain  a 
license  to  practice  architecture.  The  writer  in  the  intervals  of 
a  busy  professional  life  has  managed  to  find  time  to  teach  in  even- 
ing schools  a  deserving  class  of  men  who  entered  the  offices  of 
architects  and  contractors  at  too  early  an  age. 

Samuel  Butler  says:  "  There  are  plenty  of  things  that  most  boys 
would  give  their  ears  to  know;  these  and  these  only  are  the  proper 
things  for  them  to  sharpen  their  wits  upon."  It  happens  that  a 
great  many  boys  have  a  taste  for  drafting  and  like  to  watch  con- 
struction work.  Under  the  guidance  of  woefully  ignorant  teachers, 
lacking  practical  experience  outside  the  class  room,  drafting  is 
thought  to  be  an  end;  and  equipped  with  a  certain  facility  in 
elementary  drafting  these  unfortunate  boys  go  forth  to  seek  em- 
ployment. They  find  it,  and  after  laboring  a  few  years  discover 
that  draftsmen  who  are  merely  draftsmen  are  truly  unfortunate 
beings.  Never  enough  to  go  around  in  brisk  times,  they  are  a 
drug  on  the  labor  market  in  dull  times.  The  boys  were  given  what 
they  were  ready  to  give  their  ears  to  know,  but  their  immature 
judgment  was  at  fault  and  the  judgment  of  their  teachers  was  no 
better. 

When  a  realization  comes  of  the  fact  that  a  man  must  "  learn 
more  to  earn  more,"  Samuel  Butler  is  there  again  with  a  wise 
remark  as  follows:  "The  rule  should  be  never  to  learn  a  thing  till 
one  is  pretty  sure  one  wants  it,  or  that  one  will  want  it  before 

3 


4  PREFACE 

long  so  badly  as  not  to  be  able  to  get  on  without  it."  No  one 
knows  the  truth  of  this  remark  more  than  the  men  who  made  a 
false  start  and  got  into  the  low  paid  trade  of  drafting.  The  pity 
is  that  some  one  with  sufficient  intelligence  had  not  made  them 
understand  at  the  age  of  sixteen  that  a  draftsman  should  know 
something  more  than  just  enough  to  draw  lines  on  paper,  copying 
examples  of  the  work  of  other  men.  If  this  sort  of  knowledge 
were  pounded  into  them  they  would  "  give  their  ears  "  to  know 
enough  to  fit  them  for  advancement. 

The  realization  comes  to  few  before  the  age  of  twenty-five.  Many 
are  by  that  time  married  and  to  these  men  the  first  "  lay-off" 
comes  shortly  after  marriage,  because  periods  of  business  depression 
are  just  far  enough  apart  to  allow  this.  Few,  if  any,  have  attended 
High  School  and  few  have  graduated  from  High  School.  This  is 
the  sort  of  material  that  came  to  the  author  for  many  years  in 
his  evening  class  work.  Not  many  of  the  men  could  find  time 
outside  of  class  hours  to  work  problems.  None  of  them  were  sure 
of  themselves  when  it  came  to  working  problems.  The  men  most 
illy  prepared  to  understand  a  mathematical  demonstration  were 
most  eager  to  know  "  why."  Many  hours  were  spent  in  trying 
out  ways  to  demonstrate  truths  in  structural  mechanics  and  the 
mechanics  of  materials,  so  that  men  skilled  only  in  arithmetic 
could  understand  them.  The  men  generally  resented  a  seeming 
attempt  to  cram  them  with  formulas  and  rules  without  a  "  step 
by  step  "  explanation  of  the  work.  When  the  author  began  to 
teach  this  class  of  pupils  he  was  told  by  instructors  who  had  tried 
it  before  that  most  of  them  were  too  thick-headed  to  do  anything 
with.  The  author  found  it  otherwise.  Where  men  were  dull  it 
was  generally  because  they  were  overworked  or  were  struggling 
in  deep  financial  sloughs. 

The  men  wished  to  learn.  That  much  was  certain.  If  they  did 
not  wish  to  learn  they  could  have  had  a  better  time  elsewhere  and 
been  in  Docket  by  the  amount  of  the  fees  they  paid  for  instruc- 
tion. The  author  told  them  at  the  beginning  of  each  term  that 
if  they  failed  to  get  their  money's  worth  they  could  blame  him, 
for  he  was  there  to  teach  them  what  they  wished  earnestly  to 
learn.  They  were  a  great  inspiration  to  him  and  those  evenings 
in  the  class  room  in  an  atmosphere  of  dogged  earnestness  and 
intense  hopefulness  will  ever  remain  fragrant  in  his  memory. 
They  paid  him  for  many  hours  he  put  in  when  he  was  tired,  trying 


PREFACE  5 

to  think  of  plain  statements  of  elementary  truths.  He  has  heard 
from  a  number  of  the  men  since,  who  told  him  that  the  work  he 
did  was  an  inspiration  to  them.  The  work  was  not  undertaken 
for  the  small  amount  the  schools  could  afford  to  pay,  but  was 
done  principally  because  he  liked  the  service.  The  great  regret 
of  his  life  has  been  that  he  could  never  secure  financial  inducement 
to  enable  him  to  make  teaching  his  life  work. 

The  book  is  written  to  reach  the  men  who  cannot  attend  even- 
ing classes  in  mechanics  of  materials  and  structural  design.  It  is 
written  also  to  be  used  as  a  text  book  in  such  classes.  The  articles 
in  Building  Age  were  mimeographed  by  a  number  of  teachers  for 
use  in  manual  training  and  high  schools  and  it  is  hoped  these 
teachers  will  use  the  book  as  a  text.  The  author  has  done  his 
best  to  make  the  subject  matter  plain.  The  book  is  peculiarly 
adapted  for  the  use  of  self -tutored  men  and  the  author  would  like 
to  hear  from  such  readers,  so  that  explanations  and  statements 
they  fail  to  grasp  may  be  ironed  out  in  future  editions. 

There  is  no  royal  road  to  learning.  In  addition  to  listening  to 
lectures  or  reading  books  the  student  must  do  a  certain  amount 
of  thinking  and  reasoning.  It  is  not  enough  to  accept  a  statement 
as  true.  The  reason  why  it  is  true  must  be  understood.  The  bur- 
den on  memory  is  lessened  as  reasons  are  grasped.  For  a  man 
studying  alone  the  mistake  should  never  be  made  of  putting  an 
entire  evening  on  the  work.  Study  one  hour  each  evening,  rain 
or  shine,  and  study  hard.  A  rest  of  an  hour  is  good  and  then 
another  hour,  or  even  half  an  hour,  of  study  will  be  found  to  clear 
matters  up  wonderfully.  Early  in  the  game  start  teaching  the 
office  boy,  for  to  teach  is  one  of  the  best  ways  to  learn.  If  the  office 
boy  cannot  be  interested  then  the  studying  is  not  being  done  right, 
for  a  man  who  is  studying  in  the  proper  spirit  becomes  some- 
what enthusiastic  over  his  progress.  The  most  commonplace 
facts  are  wonderful.  That  is  why  the  newly  found  knowledge 
should  be  passed  on  to  the  office  boy,  for  men  who  know  will  not 
care  to  have  things  they  know  placed  before  them  as  fresh 
discoveries. 

One  considerable  difficulty  in  the  path  of  the  self-tutored  man 
is  eliminated  when  he  finds  a  ready  listener  and  pupil.  The 
tiresomeness  of  studying  alone  is  hard  to  describe  and  accounts 
for  so  many  quitting  early  in  the  work.  The  writer  usually  or- 
ganized a  class,  of  some  sort  when  he  had  to  bone  in  a  hurry  and 


6  PREFACE 

found  it  of  advantage.  Even  so,  difficulties  arise,  but  they  are 
much  like  the  troubles  the  old  man  spoke  about  on  his  death  bed 
as  having  been  very  real  when  they  came,  but  as  a  matter  of  fact 
most  of  them  never  happened.  Walter  Bagehot,  in  "  Physics  and 
Politics,"  says:  "  Everybody  who  has  studied  mathematics  knows 
how  many  shadowy  difficulties  he  seemed  to  have  before  he  un- 
derstood the  problem,  and  how  impossible  it  was,  when  once  the 
demonstration  had  flashed  upon  him,  ever  to  comprehend  those 
indistinct  difficulties  again,  or  to  call  up  the  mental  confusion 
that  admitted  them." 


CONTENTS 


CHAPTER  I  PAGE 

External  Forces 9 

Moments  —  Reactions  —  Relation  Between  Shear  and  Moment 
Diagram  —  Shear  —  Graphical  Methods  for  Moment,  Shear,  etc.  — 
Beams  Resting  on  Two  Supports  —  Bending  Moment  at  any  Point 
on  a  Beam  on  Two  Supports  —  Rule  for  Obtaining  the  Bending 
Moment ;  at  any  Section  of  a  Beam  —  General  Method  for  Position 
of  Maximum  Bending  Moment  —  General  Method  for  Locating 
Point  of  Zero .  Shear  —  Overhanging  Beams  — •  Equivalent  Dis- 
tributed Loads  —  Restrained  Beams  —  Continuous  Beams 

CHAPTER  II 

Internal  Forces      56 

Moment  of  Resistance  —  Elastic  Limit  —  Modulus  of  Elasticity  — 
Reinforced-concrete  Beams  —  Shearing  Resistance  —  Shear  in 
Wooden  Beams  —  Modulus  of  Rupture  —  Deflection  —  Deflec- 
tion Formulas  —  Aids  to  Computation 

CHAPTER  III 

Problems  in  Design  of  Beams 85 

Practical  Problems  in  Design  — •  Beams  on  a  Slope  —  Buckling  of 
Beams  —  Stiffness  of  Wood  Beams 


CHAPTER  IV 

Girders  and  Trusses 102 

Plate  Girders  —  Beams  with  Uniform  Stress  —  Trussed  Beams  — 
Trusses  —  Roof  Loads  —  The  Signs  Used  for  Stresses 

CHAPTER  V 

Joints  and  Connections 136 

Table  of  Upset  Screw  Ends  for  Round  and  Square  Bars  —  Table  of 
Weights  and  Areas  of  Square  and  Round  Bars  and  Circumferences 
of  Round  Bars  — Table  of  Tank  Iron  and  Steel,  Weight  of  Super- 
ficial Foot  —  Designing  Joints  —  Intermediate  Joints  in  Trusses 
—  Pin  Connections  —  Rivets  and  Rivetting  — Table  of  Shearing  and 
Bearing  Value  of  Rivets  for  Quiescent  Loads  as  Used  in  Build- 
ings —  Standards  —  Secondary  Stresses  in  Framed  Structures 
7 


CONTENTS 


CHAPTER  VI  PAGE 

Graphic  Statics 207 

Unequal  Loading  —  Wind  Force  —  Concentrated  Loads  —  Maxi- 
mum and  Mjnimum  and  Reversed  Stresses  —  Wind  on  a  Curved 
Roof  —  Cantilever  Trusses  —  Accuracy  m  Drawing  —  Continuous 


CHAPTER  VII 

Columns  and  Structures 241 

Radius  of  Gyration  —  Straight-line  Formula  —  To  Proportion 
Struts  or  Compression  Members  —  Eccentric  Loads  on  Columns  — 
Wind  Bracing  for  Columns  and  Frames  —  Loads  on  Columns  in 
Buildings  —  Column  Brackets  and  Bases  —  Eccentric  Loads  on 
Footings  —  Eccentric  Loads  on  Column  Base  —  Attaching  Column 
Bases  to  Footings  —  Cantilever  Footing  —  Stresses  in  Towers  — 
The  Design  of  Chimneys  —  Tanks  and  Retaining  Walls 


Index 295 


PRACTICAL 
STRUCTURAL  DESIGN 


CHAPTER  I 
External  Forces 

A  STRUCTURE  is  a  combination  of  parts  designed  to  hold 
in  equilibrium  definite  forces  and  in  this  book  the  word 
"structure"  is  limited  to  buildings. 

The  intention  being  to  give  a  modern  treatment  in  the  plainest 
possible  manner,  it  is  necessary  to  settle  upon  a  definite  termin- 
ology, —  that  is,  upon  a  system  of  shorthand  symbols  to  use  — 
for  it  is  best  to  present  rules  in  a  condensed  manner  in  order  that 
every  step  in  an  operation  may  be  quickly  apprehended.  A  rule 
thus  written  becomes  a  formula.  Formulas  are  merely  mathe- 
matical shorthand,  and  when  a  formula  is  seen  it  is  not  algebra. 
Algebra  is  useful  hi  deriving  a  formula,  but  when  the  formula  is 
presented  it  requires  only  the  use  of  common  arithmetic  to  solve  it. 
Let  W  =  a  uniformly  distributed  load. 

Let  w  =  a  unit  of  a  distributed  load.  On  a  panel  w  is  the  load 
per  square  foot.  On  a  beam  or  girder  w'is  the  load  per 
lineal  foot.  Then  it  follows  that  W  =  w  multiplied 
by  the  span. 

P  =  a    concentrated    load.      When    several    concentrated 
loads  are  used  the  different  loads  are  designated  by 
subscripts,  as  P,  —  P,,  —  P,  „  —  P,,  „  etc. 
S  =  clear  span  between  supports. 

L  =  length  of  span  used  for  obtaining  a  bending  moment. 
That  is,  a  beam  may  extend  from  one  support  to  another 
with  a  span,  S,  measured  from  face  to  face  of  sup- 
port, and  this  will  be  used  in  obtaining  W,  the  total 
load  on  the  span.  To  find  the  bending  effect  of  the 
load  it  is  necessary  to  use  a  length  measured  from 
center  to  center  of  supports,  or  from  points  back  of 


10  PRACTICAL  STRUCTURAL  DESIGN 

the  face  of  the  supports  and  this  length  is  L,  which 

is  always  greater  than  S. 
I  =  any  portion  of  L  or  S  used  to  obtain  a  distance  from 

the  support  to  the  center  of  gravity  of  a  load.    When 

distances  are  required  to  several  loads,  subscripts  cor- 
responding to  those  under  the  respective 
loads  are  used.  Thus  to  load  P,  we  use  /,; 


J   to  load  PH  we  use  L  etc. 


5. \—->\       The  use  of  the  letters  is  illustrated  in  Fig. 

— *-7 *l      I    1,  in  which  a  cantilever  beam  is  shown;  and 

in  Fig.  2,  in  which  is  shown  a  beam  on  two 

Fig.  1  —  TheCantilever    SUppOrts.     To  illustrate  a  uniformly  distri- 
buted load  the  weight  of  the  beam  is  used. 

Sometimes  the  letters  a,  6,  c,  etc.,  are  used  to  designate  portions 
of  a  span  or  lengths  less  than  the  whole,  but  these  will  be  dealt 
with  as  they  may  arise.  The  letters  x,  y,  z,  etc.,  are  used  simi- 
larly and  will  be  dealt  with  when  required. 

DEAD  LOAD.    The  weight  of  the  structure  and  permanent  loads. 

LIVE  LOAD.  The  load  the  structure  is  designed  to  carry  in  ad- 
dition to  the  dead  load.  The  live  load  consists  of  machinery, 
merchandise,  people,  etc. 

IMPACT.  The  effect  of  a  live  load  in  motion.  It  is  added  to  the 
live  load  and  varies  from  twenty-five  per  cent  for  a  slowly  moving 
load  on  a  rigid  structure  to  one 
hundred  per  cent  for  a  live  load 
suddenly  applied. 

WIND.    This  force  acts  perpendi- 
cularly to  the  pressed  surface,  so  is 
horizontal  on  the  sides  of  a  building 
and  is  a  diagonal  load  when  it  acts    Fig  2  _  Beam  Ites^ng  on  Two 
on  a  sloping  roof.  Supports 

Moments 

A  cantilever  beam  is  supported  at  one  end  and  may  be  loaded 
at  any  point  with  a  concentrated  load.  The  load  tends  to  bend 
the  beam  down.  The  uniformly  distributed  weight  of  the  beam 
tends  to  bend  the  beam  also,  so  on  all  loaded  beams  there  must  be 
considered  the  actual  weight  of  the  beam  plus  added  loads  it  may 
carry.  The  name  "  cantilever  "  gives  a  hint  that  the  bending  action 
is  similar  to  that  of  a  lever. 


EXTERNAL  FORCES  11 

All  formulas  and  rules  for  obtaining  the  effect  of  loads  on  beams 
are  derived  from  the  fundamental  principle  of  the  lever.  It  makes 
no  difference  whether  the  beam  rests  on  one,  two,  or  more  sup- 
ports, the  principle  of  the  lever,  as  represented  by  the  cantilever 
beam  in  Fig.  1,  gives  the  key  for  obtaining  the  desired  information. 

All  formulas  and  rules  for  obtaining  the  strength  of  a  beam  to 
resist  the  bending  effect  are  similarly  based  on  the  principle  of 
the  lever. 

The  bending  effect  is  termed  a  "  Bending  Moment "  and  the  resist- 
ing effect,  dependent  upon  the  form,  size,  and  material,  of  the 
beam,  is  termed  the  "  Resisting  Moment."  The  bending  moment 
is  first  found  and  then  a  beam  having  an  equal  resisting  moment 
is  used  to  carry  the  load. 

In  structural  design  the  moment  may  be  compared  to  the  com- 
mon denominator  in  problems  involving  fractions.  There  are  two 
quantities  which  must  be  reduced  to  a  common  measure  before 
operations  involving  both  can  be  performed.  This  explains  why 
engineers  invariably  equate  (make  equal)  the  bending  moment 
and  resisting  moment  instead  of  working  by  rules  derived  by  other 
men.  Each  man  who  does  much  designing  work  derives  rules 
for  himself  because  only  by  so  doing  can  he  be  certain  of  their 
correctness.  When  the  underlying  principle  of  moments  is  under- 
stood no  man  should  have  trouble  in  verifying  rules  which  he  may 
run  across  in  his  work  or  reading. 

Tables  are  published  of  resisting  moments  of  standard  size 
beams  from  which  a  designer  may  readily  obtain  a  beam  to  resist 
a  bending  moment,  which  is  calculated  for  each  case.  Spans  and 
also  loads  to  be  carried  on  the  spans  vary  considerably,  every 
building  presenting  a  number  of  different  combinations.  All 
rules  and  formulas  apply  to  beams  which  are  secured  against  side- 
wise  bending. 

When  the  resisting  moment  is  greater  than  the  bending  moment 
there  is  obtained  a  factor  of  safety. 

Let  M  =  moment.     This  may  be  either  bending  or  resisting 

moment. 

Mb  =  bending  moment.     The  subscript  is  used  only  when 
both  moments  are  used  in  the  same  expression,  and 
there  must  be  some  distinguishing  mark. 
Mr  =  resisting  moment,  the  subscript  being  used  only  when 
the  subscript  b  is  used  for  the  bending  moment. 


12  PRACTICAL  STRUCTURAL  DESIGN 

The  factor  of  safety  =  -—• 

Mb 

A  Moment  is  the  product  of  a  force  multiplied  by  the  distance 
through  which  it  acts. 

In  Fig.  1  the  load  P,  acts  through  a  distance  Lr  The  formula  is 

M  =  PL, 

Forces  act  through  the  center  of  gravity  of  bodies,  and  a  load 
is  a  force,  for  it  tends  to  bend  the  beam  down.  The  length  L,  is 
measured  from  the  center  of  the  support  to  the  center  of  gravity 
of  the  load. 

For  a  uniformly  distributed  load  the  center  of  gravity  is  at  the 
center,  which  for  the  beam  will  be  one-half  of  L,  so  the  formula 
for  the  bending  moment  due  to  the  uniformly  distributed  load  is 


The  total  moment  on  the  beam,  when  W  is  the  weight  of  the 
beam,  is  WT 

M  or  Mb  =  PL,+^ 

In  Fig.  6  a  cantilever  beam  carries  two  concentrated  loads.    For 
this  condition  wr 


For  more  than  two  loads  the  formula  will  be  the  same,  it  being 
only  necessary  to  obtain  the  moment  for  each  load  and  for  the 
weight  of  the  beam  and  add  them  together. 

When  the  load  is  in  pounds  and  the  distance  to  the  center  of 
gravity  is  in  feet  the  bending  moment  is  in  foot  pounds.  When 
the  distance  is  in  inches  the  bending  moment  is  in  inch  pounds. 
A  bending  moment  in  foot  pounds  is  converted  into  inch  pounds 
by  multiplying  by  12.  A  bending  moment  in  inch  pounds  is 
converted  into  foot  pounds  by  dividing  by  12. 

The  following  examples  will  illustrate  the  foregoing  formulas  : 

1.  A  cantilever  beam  projecting  10  ft.  beyond  a  wall  and  weigh- 
ing 50  Ibs.  per  lineal  foot  carries  a  concentrated  load  of  400  Ibs. 
at  a  point  7  ft.  from  the  wall.  Find  the  bending  moment  in  foot 
pounds. 

The  total  load  is  500  Ibs.  (weight  of  beam)  +  400  Ibs.  =  900  Ibs. 
Assume  the  beam  to  be  fastened  in  the  wall  1  ft.  and  the  center  of 


EXTERNAL  FORCES  13 

bearing  is  then  6  in.  from  the  face  of  the  wall.  The  length  used 
for  the  beam  will  be  10  ft.  6  in.  (10.5  ft.)  and  the  distance  to  the 
concentrated  load  will  be  7  ft.  6  in.  (7.5  ft.). 

Mb  =  (400  X  7.5)  +  (5°°  *  10'5)  =  5625  ft.  Ibs. 

Add  a  load  of  200  Ibs.  4  ft.  from  the  wall  and  a  load  of  300 
8  ft.  from  the  wall. 

Mb  =  (400  x  7.5)  +  (200  x  4.5)  +  (300  x  8.5)  + 


The  reader  will  notice  that  the  distance  to  the  center  of  gravity 
has  in  all  cases  been  measured  back  from  the  face  of  the  support, 
but  the  length  used  in  computing  the  weight  of  the  beam  was  the 
clear  length.  In  this  way  all  the  weights  are  those  clear  of  the 
supports,  for  the  portion  of  the  beam  resting  on  the  support  has 
no  effect  on  the  bending  moment. 

It  is  wrong  to  use  only  the  distance  from  the  face  of  the  support 
for  cantilever  beams,  or  the  clear  span  between  supports  for 
simply  supported  beams,  when  figuring  a  bending  moment,  as  it, 
throws  all  the  bearing  on  the  edge.  By  using  the  longer  distance 
in  computing  bending  moments  a  stiff  er  beam  is  secured.  Com- 
mercial designers  in  competitive  work  invariably  use  the  distance 
measured  from  the  face  of  supports,  as  they  thereby  save  a  little 
material  and  are  enabled  to  cut  down  cost.  All  designs  should  be 
prepared  by  men  who  have  no  other  interest  than  that  of  securing 
for  the  owner  a  design  which  is  not  "  skinned." 

Continuous  beams  with  uniform  moment  of  inertia  are  not 
subject  to  all  the  limitations  of  simply  supported  beams.  This 
will  be  discussed  later. 

Reactions 

The  loads  acting  downward  exert  an  action  on  the  beam, 
which  is  resisted  by  the  strength  of  the  bearing.  Considered  theo- 
retically the  bearing  exerts  an  upward  force  pushing  against  the 
downward  force  and  equal  in  amount.  This  is  an  example  of  the 
old  saying,  "  Action  and  reaction  are  always  equal  and  in  opposite 
directions." 

The  reactions  being  equal  to  the  load,  it  follows  that  the  reac- 


14 


PRACTICAL  STRUCTURAL  DESIGN 


tion  of  a  cantilever  beam  is  equal  to  the  sum  of  the  weight  of  the 
beam  and  all  loads  it  may  carry. 

R  =  reaction. 

For  beams  carried  on  two  or  more  supports  a  method  will  be 
given  later  for  computing  the  reactions  on  each  support. 

Shear 

Shear  is  a  downward  cutting  force  exerted  at  the  edge  of  each 
support.  It  is  called  shear  because  if  the  material  is  soft  the  edge 
of  the  support  will  cut  it.  A  piece  of  butter  resting  on  the  edges  of 
two  upturned  knife  blades  is  as  good  an  example  as  any,  perhaps, 
of  true  shear. 

V  =  shear.  It  is  always  equal  to  the  reaction  at  the  support, 
and  at  other  points  on  the  beam  varies  according  to  laws 
to  be  hereafter  explained. 

The  use  of  the  capital  V  to  designate  shear  may  be  explained 
by  its  resemblance  to  a  sharp  cutting 
edge.  Mathematicians  may  give  an- 
other reason,  but  the  writer  is  inter- 
ested in  fixing  a  fact  in  the  mind  of 
the  student. 


Graphical  Methods  for  Moment, 
Shear,  etc. 

In  Fig.  3  a  concentrated  load  acts 
at  the  point  D  on  the  beam,  AD. 
The  beam  is  drawn  to  any  scale  and 
the  load  shown  on  it,  as  in  the  figure. 
Underneath  is  drawn  the  line  A,D, 
and  the  bending  moment  at  the  sup- 
Plot  this  to  any 


(c) 


Shear 


Fig.  3  —  Concentrated  Load  at   port  is  computed. 

scale  (say  1000  Ibs.  =  1  in.)  on  the 

vertical  line  A,  A,,.  Connect  AH  to  D,  by  a  straight  line  AH  D,  as 
shown  at  (6).  To  find  the  bending  moment  at  any  intermediate 
point  on  the  beam  drop  a  vertical  line  across  the  diagram  and  the 
length  of  the  line  intercepted  between  the  upper  and  lower  lines 
of  the  bending  moment  triangle  gives  the  bending  moment.  For 
example  the  bending  moment  at  B  is  given  by  the  line  B,B,,  and 
the  bending  moment  at  C  is  given  by  the  line  C,C,,.  The  vertical 
lines  are  forces  and  the  horizontal  lines  are  lengths,  the  closing 


EXTERNAL  FORCES 


15 


line  of  the  triangle  being  merely  a  closing  line.    There  is  no  bend- 
ing moment  at  D. 

At  (c)  is  drawn  the  shear  diagram.  The  load  being  concen- 
trated at  the  end,  the  shear  of  course  is  constant  from  the  point 
of  application  of  the 
load  to  the  support. 
The  shear  may  be 
found  at  any  point  by 
measuring  a  vertical 
line  at  the  point  across 
the  shear  diagram. 


Relation  between 

Shear  and  Moment 

Diagram 

Assume  line  CfC,,  to 
be  dropped  across  the 
shear  diagram.  Then 
the  area  of  the  shear 
diagram  to  the  right  of 
the  line,  that  is  to  the 
free  end,  gives  the 
bending  moment  at  C. 
Similarly,  the  area  of 
the  shear  diagram  to 
the  right  of  line  B,,B, 
gives  the  bending  mo- 
ment at  B.  The  area 
is  in  foot  pounds  be- 
cause the  vertical  di- 
mension is  expressed  in 


Moment  Diagram 


Shear  Diagram 
Fig.  4  —  Several  Loads  on  a  Beam 

The  relation  is  true 


pounds  and  the  horizontal  dimension  in  feet, 
in  all  cases  and  must  not  be  forgotten. 

In  Fig.  4  is  shown  the  case  of  several  loads  on  a  beam.  The  bend- 
ing moment  at  D  =  0.  The  bending  moment  at  C  =  8  X  600,  and 
this  is  plotted  as  line  CC,.  The  bending  moment  at  B  =  (15  x  600) 
+  (7  x  400),  and  this  is  plotted  as  line  BB,.  The  bending  mo- 
ment at  A  =  (19  x  600)  +  (11  X  400)  +  (4  x  200),  and  this  is 
plotted  as  line  A  A,. 

The  shear  diagram,  at  (c),  shows  the  shear  at  the  right  end 


16 


PRACTICAL  STRUCTURAL  DESIGN 


The  shear  at  the  end 


Beam 


to  be  constant  and  equal  to  the  load  at  the  end  until  it  reaches 
the  next  load.  It  immediately  changes  to  the  sum  of  the  two 
loads  and  continues  as  a  constant  amount  to  the  next  load,  when 
it  immediately  becomes  the  sum  of  the  three  loads,  continuing 
thus  to  the  point  of  support. 

In  Fig.  5  are  shown  the  diagrams  for  a  uniformly  distributed 
load. 

0  and  the  moment  at  the  end  =  0. 
The  shear  diagram  is  a  triangle  and 
the  shear  at  intermediate  sections 
varies  as  the  vertical  depth  of  the 
triangle  at  the  respective  sections. 

The  bending  moment  diagram  is 
formed  by  a  closing  line  which  is  a 
semi-parabola.  This  may  be  proven 
by  dividing  the  uniform  load  into  any 
number  of  loads  equal  in  amount  and 
finding  the  moment,  as  in  Fig.  4,  for 
each  load  and  connecting  the  ends  of 
the  moments  drawn  to  scale.  The 
closing  line  is  a  broken  line,  which 
approaches  a  curve,  depending  upon 
the  number  of  unit  loads  into  which 
the  uniform  load  is  divided.  By  using 


Moment 
Diagram 


fc) 


Shear  Diagram 


Fig.  5  —  Diagrams  for  a  Uni- 
formly Distributed  Load 


a  sufficient  number  the  closing  line  becomes  a  parabolic  curve. 

If  we  assume  the  unit  load  to  be  one  foot  long  and  call  it  w,  and 
the  length  from  any  point  to  the  free  end  is  I  feet,  the  moment 

at  any  point  is 

,,       wl- 
Mb  =       ' 


By  using  this  formula  points  may  be  plotted  one  foot  apart  and 
the  ends  connected  by  using  a  French  curve. 

In  Fig.  6  is  illustrated  the  effect  of  concentrated  loads  combined 
with  a  uniformly  distributed  load.  When  a  beam  carries  concen- 
trated loads  the  shear  and  moment  diagrams  are  drawn  as  de- 
scribed for  the  uniform  load.  Then  the  moments  and  shears  due 
to  the  concentrated  loads  are  drawn  upward  from  the  top  hori- 
zontal line  of  the  diagrams.  The  total  moment  at  any  point  is 
the  distance,  BB,  or  CC,  from  the  top  closing  line  to  the  bot- 
tom closing  line,  the  intermediate  horizontal  line  being  disregarded. 


EXTERNAL  FORCES 


17 


(a) 


In  Fig.  7  is  shown  the  actual  shear  when  the  loads  have  a  defi- 
nite width  and  when  the  beam  rests  on  a  support  of  a  definite 
width.  At  the  face  of  the 
support  the  shear  is  a  maxi- 
mum and  it  is  zero  at  the 
end  of  the  support.  It  is  not 
usual  to  show  this  in  shear 
diagrams,  for  it  complicates 
the  drafting  work  without 
enough  benefit  to  pay  for  the 
trouble.  The  slight  difference 
increases  the  factor  of  safety. 


Beams  Resting  on  Two 
Supports 

To  determine  bending  mo- 
ments on  beams  on  two  or 
more  supports,  it  is  necessary 
to  find  first  the  amount  of 
the  reactions. 

In  Fig.  8  a  concentrated 
load  is  carried  on  a  beam 
resting  freely  on  two  supports. 
For  convenience  we  adopt  the 
conventional  method  of  be- 
ginning at  the  left  end  as  in 
reading.  The  word  "conven-  ^  ^ 
tional "  has  the  same  root  form 
as  the  word  "convenience" 
so  may  be  easily  remembered. 

Common  sense  assures   us 


(c) 


Shear  Diagram 


,,,.-,  i     •      •  Fig.  6  —  A  Cantilever  Beam  Carrying 

that    if   the    load    is   in    the  Two  Concentrated  Loads 

middle  one-half  will  be  carried 

by  each  support.  Let  us  imagine  the  concentrated  load  to  be 
stationary  and  the  ends  of  the  beam  pushed  up  by  the  reactions. 
The  reactions  being  equal  to  the  load,  there  is  no  movement,  but 
the  forces  actually  exist. 

Each  reaction  being  one-half  of  P,  then  RI  =  ^  and  R2  =  -• 

2  2 

This  gives  two  cantilever  beams  with  moments  acting  about 


18  PRACTICAL  STRUCTURAL  DESIGN 

(around)  the  load,  with  a  lever  arm  =  —     The  bending  moment 
under  the  load  (middle  of  span)  is 

,,      P       L      PL 


The  reader  is  not  to  forget  that  in  computing  reactions,  which 

take  into  account  only  the 
weight,    we   use   the    clear 
span  (5),  but  in  computing 
the    moment   we    use    the 
length  (L)  from  the  center 
of  bearing  on  the  support. 
*n  FiS-  9  a  uniform  load 
(represented  by  the  weight 
ear  Diagram  of  the  beam)  is  carried  Qn 

.  7  —  A  Cantilever  Beam  Supporting     two   supports   on    which   it 
an  I-beam  rests  freely. 

w 

Each  reaction  =  •_->  in  which  W  =  wS. 

This  gives  two  cantilevers  with  moments  acting  about  the  center 
of  gravity  of  the  beam,  which  being  in  the  middle  of  the  span 

makes  the  length  of  the  moment  arms  =  j-  measured  from  the 

middle  of  the  span  to  the  center  of  gravity  of  each  half  span. 
The  bending  moment  is  ^  ...................  ^  ___  ...............  « 

W      L      WL        wSL  "  *        /t\ 

M  =  -^  X  -T  =  -f>  or  -£-•       I  _  (£2  _  I 

A  4  O  O  "^  W&- 


w 

The  load  -^  acts  as  a  uni-    Fig   8  _  Concentrated  Load  at  Mid. 
formly    distributed    load   and        ^n  with  Beam  on  Two  Supports 
not  as  a  concentrated  load  at  the  end,  although  it  is  equal  to  the 
reaction. 

The  reader  will  notice  that  the  multiplication  sign  (x)  is  not 
used  between  letters,  for  when  several  letters  are  written  together 
in  formulas  it  means  they  are  to  be  multiplied  together.  If  the 
multiplication  sign  were  used  it  might  be  mistaken  for  the  letter 
x  when  written.  The  multiplication  sign  is  always  used  between 
figures.  Thus,  WL  stands  for  W  x  L,  but  67  means  sixty-seven 
and  does  not  mean  6x7=  42. 


EXTERNAL  FORCES  19 

In  Fig.  10  a  concentrated  load  is  assumed  to  be  at  the  middle 
of  an  opening  spanned  by  a  beam  of  uniform  weight.  The  reac- 

p       w      P  +  W 

tions  are  as  follows:  R\  =  R2  =  -^  +  -^-  =  — ~ The  bending 

Zi  6  A 

,,      PL      WL 

moment  is  M  =  —j-  -\ — ^-. 

In  Fig.  11  several  concentrated  loads  are  shown  on  a  span  and 
the  reactions  are  to  be  _^ 

found. 

Commencing  at  the  left 

end   take   moments    about        1 1  ely          ^         c\g. 

RI.     Thus  the  bending  mo. 
ment  at  the  left  support  is 

T\f  _  pn    i    p  n    _i_  p  n          Fig.  9  —  Uniform  Load  on  Beam  Resting 

+  /V*,  +  n<V  on  Two  Supports 

This  moment  will  have  a 

tendency  to  carry  the  far  end  of  the  beam,  at   #2,  downward 
unless  a  supporting  force  is  exerted  to  hold  it  in  position. 

Here  the  principle  of  moments  is  again  applied.  The  moment 
of  a  force  is  equal  to  the  force  multiplied  by  the  distance,  or  arm, 
through  which  it  acts,  so  it  is  necessary  to  have  at  R2  an  upward 
moment  equal  to  the  downward  moment.  This  is  obtained  by 
dividing  the  downward  moment  by  the  span  length. 

F---FF^ "i    «2=po+prp"a" 

" ~*"7r\  #1  =  (sum  of  the  loads)  —  #2. 

(  h*    I 

pj ^ ^~^- 1-|      This  is  proved  when  we  con- 
sider that  the  sum  of  the  re- 
actions is  equal  to  the  total 
Fig.  10  — Concentrated  Center  Load  and  load.      The    amount    of    each 
Uniform  Load  on  Beam  Resting  on       reaction   may  be   checked  by 

taking  moments  from  the  right 
end  instead  of  the  left  and  working  as  before. 
Example:    Let  a    =    3  feet  and  P    =  200  Ibs. 
a,   =    7    "      "    P,  -  300   " 
a,,  =  11    "      "    P,,  =  250   " 
Span  =  15  feet.  750  =  total  load. 

7?        (3  X  200)  +  (7  x  300)  +  (11  X  250) 

/ta  = T^ =  363.33  Ibs. 

#1  =  750  -  363.33  =  386.67  Ibs. 


20  PRACTICAL  STRUCTURAL  DESIGN 

Checking  by  taking  moments  about  the  right  end. 
(4  x  250)  +  (8  X  300)  +  (12  x  200) 


d»b.b7 


R2  =  750  =  386.67  =  363.33  Ibs. 


The  same  results  may  be  obtained  by  common  proportion,  but 
upon  examination  the  method  by  moments  is  seen  to  be  exactly 

_a^      the  same  thing,  but  with 

t- *--•--- — »j  less  work.   If  the  reactions 

°  '^\       /^\       /E\  were  found  by  the  common 

school  arithmetical  process 
of    proportion    the    work 

would  be  longer  and  mis- 
Fig.   11 -Several   Concentrated  Loads  on    takeg  more        t   to  occur 
Beam  Having  Two  Supports  „,,  , .     .      . 

The  method  of  moments 

is  the  shortest  and  neatest  way  of  working. 

In  Fig.  12  three  concentrated  loads  are  shown  on  a  beam  of 
which  the  weight  is  uniformly  distributed.  First  find  the  reactions 
due  to  the  concentrated  loads.  Then  add  to  each  reaction  half 
the  weight  of  the  uniformly  distributed  load. 

Example.  —  Assume  a  beam  having  a  weight  of  50  Ibs.  per 
lineal  foot  on  a  15-ft.  span,  carrying  the  concentrated  loads  given 
in  the  last  example.  What  are  the  reactions? 

Answer.  —  Weight  of  beam  =  15  x  50  =  750  Ibs. 
The  reaction  at  each  end  =  0 .„ 

-=-  =  375  Ibs. 


761. 67  Ibs. 
#2=363.33+375=   738.33  " 

Total  load        1500.00  " 

When  the  bending  moment    Fig.  12  —  Uniform  Load  and  Several 
is  wanted  at  any  section  of  a  Concentrated  Loads  on  Beam 

beam  we  assume  the  beam  at  Havmg  Tw°  Supports 

the  section  to  be  supported  at  the  section  and  moments  are  taken 
about  it  as  if  it  were  a  cantilever  beam.  The  reaction  is  an  up- 
ward force  creating  an  upward  moment  and  the  loads  are  down- 
ward forces  creating  a  downward  moment.  The  difference  between 
the  moments  in  opposite  directions  is  the  bending  moment  at  the 
chosen  section.  When  a  beam  is  freely  supported  on  two  or  more 


EXTERNAL  FORCES 


21 


supports,  this  is  always  a  positive  (downward)  moment,  the  beam 
being  in  tension  on  the  lower  side  and  in  compression  on  the 
upper  side  between  supports. 

Bending  Moment  at  any  Point  on  a  Beam  on  Two  Supports 

The  loads  are  shown  in  position  and  amount  in  Fig.  13.    First 
find  the  reactions. 


751.56 


K 9>—  --— H 

|^ —13.5' 


Fig.  13  —  Example  of  Uniform  Load  and  Several  Concentrated  Loads  on 
Beam  on  Two  Supports 


2  =  (4.5x200)  +  (7.5  X30Q)  +  (11.5x250)  +  (8x750) 
16 

!  =  (200  +  300  +  250  +  750)  -  751.56  =  748.44  Ibs. 

Check  for  Ri  : 

(4.5x250)  +  (8.5x300)  +  (11.5x200)  +  (8x750) 


In  finding  the  reactions  the  weight  of  the  beam  =  15  X  50  =  750 
Ibs.  This  was  multiplied  by  half  the  length  of  the  span,  which 
gave  the  quantity  above,  8  X  750. 

What  is  the  bending  moment  at  the  section  zz? 

The  section  xx  is  2  ft.  from  the  face  of  the  left  support. 
The  span  face  to  face  of  supports  =  15  ft.,  but  the  length  center 
to  center  of  bearings  =  16  ft.,  assuming  a  bearing  1  ft.  long  on 
each  support.  Therefore  the  moment  arm  from  the  section  to 
the  reaction  at  the  right  end  =  13.5  ft.  The  moment  arms  to  the 
loads  from  the  section  are  marked  on  the  figure.  There  is  one 
moment  arm,  however,  of  6.5  ft.  to  be  explained.  It  is  the  length 
from  the  section  to  the  center  of  gravity  of  that  portion  of  the 
beam  lying  between  the  section  and  the  support  at  the  right  end. 
The  total  clear  span  =  15  ft.  and  the  section  is  2  ft.  from  the  left 


22 


PRACTICAL  STRUCTURAL  DESIGN 


support.    The  length  of  the  beam  therefore  to  the  right  support 
=  13  ft.,  and  one-half  =  6.5  ft.    The  weight  =  13  x  50  =  650  Ibs. 

M  =  (13.5x751.56)-[(2x200)+(5x300)+(6.5x650)  +  (9x250)]  = 

1771.06  ft.  Ibs. 

In  Fig.  14  the  section  xx  is  taken  5  ft.  from  the  left  sup- 
port.   This  leaves  the  load  of  200  Ibs.  to  the  left  of  the  section, 


(fe    @ 


=v 


RI -151.56 


-10.5 


Fig.  14  —  Another  Example  of  Uniform  Load  and  Several  Concentrated 
Loads  on  Beam  on  Two  Supports 

so  it  is  omitted  from  the  calculations.  The  load  was  used  in  ob- 
taining the  reactions,  but  in  this  present  example  it  will  be  noticed 
that  the  moment  arm  from  the  section  is  only  10.5  ft.  to  the 
reaction.  The  beam  length  =  15  —  5  =  10  ft.  and  the  weight 
=  10  X  50  =  500  Ibs.,  with  a  moment  arm  to  the  center  of  gravity 
=  5ft. 
M  =  (10.5x751.56)-[(2x300)+(5x500)+(6x250)>3291.38ft.lbs. 


0( 300\  (  7SO\ 

\^  y\      vLx 

i  \c\g. 


~j 

t~£ 

ij 

'//////////// 

\  *'*' 

K- 

Fig.  15  —  Still  Another  Example  of  Uniform  Load  and  Several 
Concentrated  Loads  on  Beam  on  Two  Supports 

In  Fig.  15  the  section  xx,  is  taken  8  ft.  from  the  left  sup- 
port. The  beam  length  =  15  -  8  =  7ft.  and  the  weight  =  7  X  50 
=  350  Ibs.,  with  a  moment  arm  of  3.75  ft.  to  the  center  of  gravity. 
M  =  (7.5  X  751.56)  -  [(3  x  250)  +  (3.75  X  350)]  = 

3574.2  ft.  Ibs. 


EXTERNAL  FORCES  23 

The  method  given  for  obtaining  the  bending  moment  at  any 
section  is  used  for  any  number  of  loads,  three  loads  being  used  in 
the  examples  for  the  sake  of  clearness.  The  computations  being 
illustrative  the  reactions  have  been  given  to  fractions  of  a  pound 
and  the  moments  have  been  given  to  fractions  of  foot  pounds. 
In  actual  work,  it  is  generally  considered  that  the  nearest  unit 
is  sufficiently  exact. 

Rule  for  Obtaining  the  Bending  Moment;  at  any  Section 
of  a  Beam 

Multiply  one  end  reaction  by  the  length  from  it  to  the  section. 
From  the  moment  thus  obtained  subtract  the  sum  of  the  moments 
of  the  loads  lying  Between  the  section  and  the  chosen  reaction,  using 
as  moment  arms  the  length  in  each  case  from  the  center  of  gravity 

of  the  load  to  the  section.     The  a H^....    _  ^ 

portion   of  the   beam   included  '^ 

between  the  section  and  the  re- 
action is  to  be  counted  as  a  load. 

A  floor  is  merely  a  shallow    Fig.  16  —  Concentrated  Load  at  Any 
beam,  usually  with  a  width  of         Point  on  Beam  on  Two  Supports 
..  ~  .     ,  as  bnown 

12  inches. 

A  beam  is  a  secondary  girder  and  the  load  is  usually  uniformly 
distributed. 

A  girder  is  uniformly  loaded  when  it  carries  the  floor  slab 
directly  without  the  floor  load  going  first  to  beams.  When  the 
floor  rests  on  beams  the  reactions  at  the  ends  of  the  beams  are 
concentrated  loads  going  to  the  girders. 

A  girder  is  generally  carried  on  walls  or  columns  and  beams  are 
generally  carried  on  girders.  A  rafter  is  a  girder  and  purlins  are 
beams,  or  joists. 

For  a  load  concentrated  at  any  point,  referring  to  Fig.  16, 

,,      Pab  ,     .      ,      . 
M  =  -^—>  for  load  only. 
Li 

The  derivation  of  the  formula  is  as  follows: 

R2  =  _£,  and  Ri  =  P  -  R2, 
Li 

Tb*-M -**..*'.*£&> 


24  PRACTICAL  STRUCTURAL  DESIGN 

The  maximum  bending  moment  for  a  single  load  is  always  under 
the  load.  If  the  weight  of  the  beam  is  included  the  moment  under 
the  load  is 

M  =  RJ)  —  —-> 

in  which  w  =  weight  of  beam  per  lineal  foot. 

Example.  —  A  beam  with  a  length  of  15  ft.  between  centers  of 
supports  carries  a  load  of  600  lbs.  at  a  point  5  ft.  from  the  left 
support.  The  weight  of  the  beam  per  lineal  foot  is  9  lbs.  What 
is  the  bending  moment? 

First.  —  For  load  only. 

,,      Pab      600  X  5  X  10      OAAA  .,   ., 

M  =  -j—  =  — =  2000  ft.  lbs. 

Li  lo 

Second.  —  For  load  and  beam. 

Pa      wL      600  X  5      135 
Rz  -  -j-  +  ~n~  =  — J-R- 1 — o~  =  *o7.5  lbs. 

...1,2  /q  v  102\ 

M  =  RJ>  -  ^-  =  (267.5  x  10)  -  f     *2     J  =  2675  -  450 

=  2225  ft.  lbs. 

Referring  to  Fig.  17,  the  load  P  acts  through  the  center  of 

gravity,  but  the  effect  is 
lessened  because  the  load  is 
distributed  over  a  portion  of 
the  beam  instead  of  acting 
at  a  point  as  it  would  were 
the  load  round  like  a  ball. 

Fig.  17  -  Partially  Distributed  Load  on  a         Consider  the  load  of  QQQ 

Beam  Resting  on  Two  Supports  as  Shown 

lbs.  in  the  last  example  to 

be  distributed  over  a  length  of  3  ft.    What  is  the  total  bending 
moment  under  the  load? 

tf.ftj-f-l 

Writing  this  out  it  appears 

q  v  102       fiOft  v  3 
M  =  (267.5  x  10)  -  ^-^        ^^  =  2675  -  450  -  225 

=  2000  ft.  lbs. 


EXTERNAL  FORCES  25 

The  formulas  for  the  reactions  are  the  same  as  though  the  load 
was  concentrated  at  a  point. 

For  two  equidistant  loads  on  a  beam,  as  in  Fig.  18,  the  formula 
for  the  reactions  reads 


The  moment  due  to  the  loads  only  is  a  maximum  under  each 
load  and  is  constant  at  all  points  on  the  beam  between  the  loads. 
M  (  for  loads  only)  =  Pa. 

This  can  be  proved.  The  loads  being  placed  on  the  beam  at 
the  same  distance  from  the  end,  call  each  load  P.  Then  the  total 
load  =  2P.  One-half  of  the 

total  load  goes  to  each  end,  /^K  /j^\ 

so  the  reaction  must  be       r|  -  —  ^  - 


—  =  P.    The    moment    is 

2  f—  —  -  L 

equal  to  the  reaction  multi-    Fig.  18  —  Two  Equidistant  Concentrated 
plied   by  the   arm   through    Loads  on  Beam  on  Two  Supports  as  Shown 
which  it  acts,  therefore  M  =  Pa. 
Adding  the  weight  of  the  beam  gives  us  under  the  load, 

M  =  R,a  -  yf 

in  which  w  =  weight  per  lineal  foot  of  beam 

a  =  length  of  beam  between  load  and  reaction. 
To  prove  that  the  moment  is  the  same  under  the  middle  of  the 
beam  as  it  is  under  each  load  we  must  multiply  the  reaction  by 
half  the  span  and  subtract  from  it  the  load  multiplied  by  half  the 
span  minus  the  length  of  the  arm  from  the  load  to  the  reaction. 
The  full  written-out  expression  is  as  follows: 

11.  P  X  \- 

NOTE.  —  In  the  above  expression  algebra  has  been  used  for  the 
first  time.  The  product  of  two  positive  (+)  quantities  is  positive. 
The  product  of  two  negative  (-)  quantities  is  positive.  The 
product  of  a  positive  (+)  and  a  negative  (-)  quantity  is  negative. 
In  the  above  expression,  where  the  subtraction  is  indicated,  the 
load  P  is  negative  and  the  first  quantity  within  the  parenthesis 
is  positive,  for  when  no  sign  is  written  the  positive  (+)  sign  is 


26  PRACTICAL  STRUCTURAL  DESIGN 

understood.  The  second  quantity  is  negative.  By  clearing 
away  the  parenthesis  and  multiplying  each  quantity  within  by 

PT  PT 

the  load  we  obtain  +  -^-  and  — — ,  which  of  course  cancel  each 

z  z 

other,  leaving  Pa,  for  P  =  reaction  when  weight  of  beam  is 
neglected. 

Example  (Fig.  19). — A  beam  15  ft.  long,  weighing  9  Ibs.  per 
lineal  foot,  carries  two  loads  of  300  Ibs.  each  at  points  distant  5  ft. 

£,_ 5, from  the  ends.  Find  the 

boh  fod\  bending  moment. 


i  [  N|/        i        NT  cgr.  I  i         First. — Under    each 


15'- 


load  for  the  loads  only. 
M  =  Pa  =  300  x  5  = 


1500  ft.  Ibs. 
Fig.  19  -Diagram  Illustrating  the  Example         Thig    moment'   ig   con. 

stant  for  each  point  on  the  beam  between  the  loads. 

Second.  —  Under  each  load  for  the  load  and  the  weight  of  the 
beam. 

R,  =  Rz  =  p  +  E=  300  +  ^-4^  =  367.5  Ibs. 
z  z 

M  =  (367.5  x  5)  -    ~-^    =  1725  ft.  Ibs. 


This  moment  is  not  constant,  for  the  weight  of  the  beam  between 
the  loads  must  be  considered.  The  increase  is  slight  and  usually 
is  negligible,  except  in  the  case  of  concrete  beams,  in  which  the 
dead  load  often  equals  or  exceeds  the  live  load.  This  example 
will  be  worked  out  in  detail,  there  being  four  distinct  steps. 

First.  —  Total  moment  at  middle  of  beam  =  Rz  X  -~- 
Second.  —  Moment  due  to  weight  of   half    the  beam.     The 
weight  =  -~—    Multiply  this  by  the  distance  from  the  middle  of 

the  span  to  the  center  of  gravity  of  half  the  beam  =  —     This 

reduces  to  -^-  X  -7  =  —  Q-,  to  be  subtracted  from  the  total  moment, 
2i          4          o 

which  took  into  consideration  the  weight  of  the  beam.    This  gives 


EXTERNAL  FORCES  27 

Third.  —  The   moment  due  to  the  concentrated  load  must 
be  subtracted.     The  load  is  multiplied  by  the  distance  from 

the  middle  of  the  beam  to  the  load  =(~n  —  a  )  J    this  giving  P  x 
I—  -  a)  =  —£-  +  Pa,  and  as  this  is  to  be  subtracted  the  minus 

sign  is  placed  before  -^~- 

z 

Fourth.  —  The  whole  expression  now  appears, 


When  adding  positive  and  negative  quantities  add  each  kind 
separately.  Take  the  difference  of  the  sums  and  prefix  the  sign 
of  the  greater  sum.  By  inclosing  the  two  negative  quantities  in 
a  parenthesis  the  sign  of  the  second  is  changed,  so  the  full  expression 
may  be  written 

PL      wU\ 


The  arithmetical  work  is  simple,  as  here  shown 

=  (300  X  5  +  367.5  x  7.5)  -  (300  x  7.5  +  9  X?25    =  1753  ft.  Ibs. 


Graphical  Methods 

To  divide  a  line  into  any  number  of  equal  parts  use  the  geo- 
metrical principle  of  similar  triangles. 


«" 3 2 3 5 5 6 7          V 

Fig.  20  —  Division  of  Line  into  Equal  Parts 

In  Fig.  20  let  AC  represent  a  line  that  is  to  be  divided  into  any 
number  of  equal  parts.  The  length  is  such  that  no  regular  scale 
can  be  used  readily  for  the  purpose.  Set  off  from  one  end  a  line, 
A  B,  at  any  angle.  This  line  is  drawn  to  some  scale  and  divided 
into  as  many  equal  parts  as  it  is  desired  to  divide  the  line  AC. 
Connect  B  to  C  and  through  the  divisions  on  the  line  AB  draw 
lines  parallel  to  BC,  as  shown. 


28 


PRACTICAL  STRUCTURAL  DESIGN 


A  number  of  curves  are  used  by  engineers  and  scientists,  the 
most  useful  of  which,  and  the  only  one  used  by  structural  de- 
signers, is  the  parabola.  This  curve  is  formed  by  cutting  a  section 


Fig.  21  —  Section  of  Cone  Developed  into  Parabola 

through  a  cone  parallel  to  the  slope,  as  shown  by  the  line  EF 
in  Fig.  21.  The  major  axis  in  all  curves  is  designated  by  the  letter 
A  and  the  minor  axis,  perpendicular  to  the  major  axis,  by  the  small 
letter  a.  The  axes  of  a  parabola  are  infinite,  so  for  this  curve  A 
designates  height  and  a  one-half  the  base. 


r 


L_ 


Fig.  22  —  The  Parabola  and  its  Equations 

In  Fig.  22  let  X  and  x  =  the  abscissas  and  Y  and  y  the  ordinates 
of  the  parabola.  The  abscissas  are  parallel  to  and  the  ordinates 
are  perpendicular  to  the  major  axis.  Assuming  X  and  Y  each 


EXTERNAL  FORCES  29 

with  a  value  of  1,  and  X  divided  into  10  equal  parts,  the  following 
values  are  obtained,  numbering  the  spaces  from  the  bottom  up. 

Points  on  X  Ordinates  (Y) 

1  0.949 

2  0.894 

3  0.837 

4  0.775 

5  0.707 

6  0.633 

7  0.548 

8  0.447 

9 0.317 

Assuming  X  and  Y  each  with  a  value  of  1,  with  X  divided 
into  8  equal  parts,  the  following  values  are  obtained : 

Points  on  X  Ordinates  (Y) 

1   0.936 

2  0.866 

3  :... 0.791 

4  0.707 

5  0.612 

6  0.50 

7  0.353 

To  construct  a  parabola  by  using  a  table  of  ordinates  erect  a 
perpendicular  at  the  middle  of  the  span  having  a  height  equal 
to  the  bending  moment,  using  any  convenient  scale.  Divide  the 
line  into  8  or  10  equal  parts  and  through  the  division  points  draw 
horizontal  lines  parallel  with  the  beam.  Multiply  the  half  span 
by  the  value  of  the  ordinate  for  any  line  and  set  off  to  scale  the 
length  of  the  ordinate.  Connect  the  ends  by  means  of  a  French 
curve  and  thus  obtain  the  parabola.  In  this  method  the  scale  for 
all  horizontal  lines  is  the  scale  used  in  drawing  the  beam. 

In  Fig.  23  another  method  is  shown.  Divide  the  span  of  the 
beam  into  any  number  of  equal  parts,  numbering  from  each  end 
as  shown.  Multiply  the  maximum  moment  by  the  product  of 
the  two  figures  under  any  line  and  divide  by  the  product  of  the 
two  equal  figures  under  the  maximum  moment  line.  The  result 
is  the  length  of  the  perpendicular  at  the  two  numbers.  Connect- 
ing the  upper  ends  of  the  perpendiculars  by  using  a  French  curve, 
the  parabola  is  drawn.  The  scale  used  in  drawing  the  perpendicu- 
lar lines  is  the  scale  used  in  setting  off  the  value  of  the  bending 
moment  at  the  middle  of  the  span. 


30  PRACTICAL  STRUCTURAL  DESIGN 

Compute  the  moments  at  the  points  shown  in 

1000  x  7  x  3 


Example. 
Fig.  23. 

1000  X  6  x  4 


5x5 

1000  X  8  x  2 
5x5 


=  960 


640 


5x5 

1000  x  9  x  1 
5x5 


=  840 


=  360 


The  parabola  in  Fig.  23  was  constructed  by  using  perpendiculars 
computed  as  shown  and  the  curve  drawn  with  a  French  curve. 

The  middle  perpendi- 
cular line  was  divided 
into  10  equal  parts 
and  the  ordinates  to 
the  major  axis  mea- 
sured off  on  the  hori- 
zontal lines  to  check 
the  accuracy  of  the 
curve.  This  is  recom- 
mended as  an  exercise 
for  the  student. 

A  graphical  method 
for  constructing  a  pa- 
rabola is  shown  in  Fig. 
24.  The  perpendicular 
representing  the  bend- 
ing   moment    at   the 
Fig.  23  —  Ordinate  Method  for  Constructing     middle  of  the  beam  is 
Parabola  , 

set  up  and  a  rectangle 

drawn,  with  a  height  equal  to  the  bending  moment  and  a  width 
equal  to  the  span.  The  horizontal  lines  of  the  rectangle  are 
divided  into  any  number  of  equal  parts  and  the  vertical  end  lines 
into  half  this  number.  In  the  example  the  horizontal  lines  are 
divided  into  8  parts  and  the  vertical  lines  into  4  parts.  From  the 
apex  radiating  lines  are  drawn  to  the  end  divisions  and  vertical  lines 
are  drawn  through  the  horizontal  divisions.  A  curve  is  drawn 
through  the  intersections  of  the  radiating  and  vertical  lines. 

All  the  common  methods  in  use  for  drawing  parabolas  have 
been  given  in  order  that  the  students  may  have  a  choice  of  methods 
as  well  as  to  show  that  in  even  the  most  ordinary  matters  there 
are  several  ways  of  accomplishing  a  result.  The  man  who  knows 


EXTERNAL  FORCES 


31 


only  one  way  for  doing  anything  is  apt  to  be  dogmatic  and  think 
his  way  alone  is  right.    An  ingenious  man  will  discover  for  himself 


Fig.  24  —  Graphical  Method  for  Constructing  Parabola 

many  ways  to  shorten  his  work  when  he  knows  he  is  not  compelled 
to  adhere  to  some  method  he  has  been  shown. 

When  the  height  of  the  parabola  is  greater  than  the  span  more 
accurate  results  may  be  obtained  than  when  the  height  and  span 
are  nearly  equal,  or  when  the  height  is  less  than  the  span.  Use 
a  scale  that  will  give  results  with  the  accuracy  required  for  the 
work.  In  some  work  the  height  of  the  parabola  may  be  much 


Fig.  25  —  Graphical  Solution  for  One  Concentrated  Load 
Resting  on  a  Beam 

less  than  the  span  and  still  give  results  sufficiently  accurate. 
When  the  height  of  the  parabola  is  not  greater  than  one-tenth 


32 


PRACTICAL  STRUCTURAL  DESIGN 


the  span,  a  semicircle  may  be  used.  For  cantilever  beams  the 
vertex  of  the  parabola  is  at  the  support.  For  beams  on  two  or 
more  supports  the  vertex  of  the  parabola  between  the  supports 
is  at  the  top. 

For  one  concentrated  load  in  the  middle  of  a  beam  resting  freely 
on  two  supports  drop  a  perpendicular  line  under  the  load  to  repre- 
sent the  amount  of  the 

a  >K- bi—          ^        bending   moment,    as 

. J^-b    4fe—  a  — M 


Fig.  26  —  Graphical  Method  for  Two  Concen- 
trated Loads  Resting  on  a  Beam 


shown  in  Fig.  25.  Con- 
nect the  lower  end,  C, 
of  the  line  by  straight 
lines  to  A  and  B. 
Measuring  vertically 
from  any  point  on  the 
center  line  of  the  beam 
to  the  bounding  line, 
ACS,  gives  the  bend- 
ing moment  at  that 
point.  The  moment 
at  any  point  may  be 
computed  also  if  the 


arithmetical  method  is  preferred  to  the  graphical  method.     The 
load  being  concentrated  at  the  middle  of  the  span, 

PL 

4  ' 


M 


To  compute  the  bending  moment  at  any  section,  xx,  distant 
y  from  the  end,  use  the  principle  of  similar  triangles 

L   PL 


from  which  (for  the  load  only) 
PL 


XL 


Py 
2' 


The  shear  diagram  is  shown  below.  For  a  concentrated  load 
the  shear  is  constant  from  the  load  to  each  end  and  the  end  shear 
is  always  equal  to  the  reaction. 

In  Fig.  26  two  concentrated  loads  are  shown.    Call  the  length 


EXTERNAL  FORCES  33 

from  either  end  a  and  that  from  the  other  end  6.    Then  under 
each  load  (for  the  load  only) 


Through  each  load  drop  a  perpendicular  and  set  off  the  bending 
moment  on   each  line 
for  the  load  above  '      ^ 

that  line.  Connect  the  T  _""  "".7".."_SM "~ 
ends  of  the  lines  to  the 
ends  of  the  center  line 
of  the  beam,  thus  mak- 
ing two  triangles.  Un- 
der each  load  is  the 
moment  due  to  the 
load,  plus  the  moment 
due  to  the  other  load 
shown  by  the  inter- 
cepts, Pb  and  PI&I. 
From  a  set  off  ac  =  Pb 
and  from  at  set  off 
aiCi  =  PI&I.  The  total 
moment  under  P  =  PC 
and  under  Pi  =  PiCi. 
Connect  the  points  by 
the  lines  acciB,  thus 
forming  a  bending  mo- 
ment diagram  (for  the 
loads  only).  The  bend- 
ing moment  at  any 
point  on  the  beam  is 
obtained  by  measuring 
from  the  center  line 


Fig.  27  —  Graphical  Method  for  Several  Loads 
on  a  Beam 


(AB),  of  the  beam  to  the  bounding  line  of  the  bending  moment 
curve. 

In  Fig.  27  is  illustrated  the  application  of  the  method  just 
described  to  three  loads.  Any  number  of  loads  may  be  similarly 
treated,  no  matter  how  unequal  in  weight  nor  how  unevenly  spaced. 

Assume  any  number  of  equal  loads  equally  spaced  as  in  Fig.  28. 
This  amounts  to  a  uniform  load,  and,  triangles  being  drawn  as 
shown,  for  each  load,  a  bending  moment  diagram  is  formed  of  which 


34 


PRACTICAL  STRUCTURAL  DESIGN 


the  boundary  is  a  parabola.    Knowing  this  to  be  true  a  parabola 
with  a  height  represented  by 

V-  =  WL 

8     ~    8 


M 


is  drawn  to  represent  the  bending  moment  for  uniformly  loaded 

beams.  The  bending 
moment  at  any  point 
is  obtained  by  measur- 
ing vertically  from  the 
center  line  of  the  beam 
to  the  parabola.  The 
bending  moment  at  any 
point  of  a  uniformly 
loaded  beam  is  com- 
puted as  follows,  call- 
ing the  distance  from 
the  end  of  the  beam  to 
the  point  in  question  x  : 

T.,       wlx      wx2 


A  moving  load,  for 
example  a  single  wheel, 
occupies  each  point  on 
the  beam  as  it  travels. 
If  a  triangle  is  drawn 
for  each  position  of  the 
load  and  the  triangles 
are  connected  by  a 
bounding  line  a  para- 
bola will  be  formed 
with  a  middle  ordinate 
equal  to 


Fig.  28  —  Graphical  Method  for  Uniformly 
Distributed  Load  on  a  Beam 

M        PL 

M  =  --> 


for  the  continuous  line  across  the  span,  of  units  each  equal  to  the 
load  on  the  wheel,  compares  with  a  uniform  load.  Consequently, 
to  ascertain  the  bending  moment  at  any  point  on  a  beam  due  to 
a  single  traveling  load,  construct  a  parabola  with  a  middle  ordinate 


EXTERNAL  FORCES 


35 


as  above  and  measure  the  ordinate  to  the  curve  at  the  point 
where  the  desired  bending  moment  is  to  be  found. 

In  Fig.  29  is  shown  the  effect  of  concentrated  loads  plus  the 
uniform  load  due  to  the  weight  of  the  beam.    The  moment  due  to 
the  uniform  weight  of  the  beam  is  computed  and  a  center  line 
measured  upward  to 
represent  this  mo- 
ment.     Construct    a 
parabola. 

Below    the    center 

line  of  the  beam  con-  i    /       x-x  , 

struct  a  moment  dia-  L/        V ?)  \ 

gram  representing  the 
effects  of  the  concen- 
trated  loads.  The 
bending  moment  at 
any  point  is  shown 
by  the  line  intercept- 
ed by  the  upper  and 
lower  boundaries  of 
the  moment  curves. 
For  example,  at  a  dis- 
tance y  from  the  left 
end  of  the  beam  the 
bending  moment  is 
shown  by  the  length 
of  the  line  xxi. 

In  Fig.  29  the  re- 
actions are  drawn  to 
scale  so  that  ac  and  bd 

each  represent  one-half  the  uniform  load.  Draw  the  horizontal 
line  ab  and  connect  c  to  d.  The  two  triangles  represent  the 
shear  to  scale  at  all  points  on  the  beam.  The  horizontal  measure- 
ments are  lengths  and  the  vertical  measurements  are  loads.  Since 
beams  usually  weigh  much  less  than  any  load  they  carry  it  is  com- 
monly stated  that  the  maximum  moment  and  zero  shear,  or  point 
where  shear  passes  through  zero,  occur  always  under  a  concen- 
trated load.  The  student  can  see  that  this  statement  must  be 
qualified  when  the  uniform  load  is  considerable.  The  combined 
shear  diagram  in  Fig.  29  is  obtained  by  adding  the  end  reactions 


Combined  Shear 


Fig.  29  —  Graphical  Method  for  Combination  of 

Concentrated  and  Uniformly  Distributed 

Loads  on  a  Beam 


36  PRACTICAL  STRUCTURAL  DESIGN 

due  to  the  different  load  systems  and  making  the  sloped  lines 
parallel  to  those  in  the  diagram  of  uniform  load  shear. 

General  Method  for  Position  of  Maximum  Bending  Moment 

First .  —  Find  the  reactions. 

Second.  —  Starting  from  either  end  add  the  loads  until  a  point 
is  reached  where  the  sum  of  the  loads  equals  or  exceeds  the  reaction 
at  that  end.  This  is  the  point  of  maximum  bending  moment. 

General  Method  for  Locating  Point  of  Zero  Shear 

First.  —  Call  the  left  reaction  positive  (+)  and  the  right  reaction 
negative  (— ). 

Second.  —  Call  each  load  negative  (— )  and  successively  subtract 
from  the  left  reaction  each  load,  prefixing  the  proper  sign  until  the 
sign  of  the  sum  of  the  quantities  changes  from  positive  (+)  to  nega- 
tive (— ).  This  is  the  point  of  zero  shear  and  maximum  bending 
moment. 

An  inspection  will  show  the  two  rules  to  be  identical. 

For  a  uniform  load  the  shear  at  any  point  distant  x  from  either 
support  is  found  as  follows : 

wL 

Shear  at  x  =  -^ —  wx. 

When  a  moving  uniform  load  is  passing  over  the  beam,  a  train 
of  small  trucks,  for  example,  the  maximum  shear  at  any  point 


When  the  load  covers  the  span  x  =  0  and  the  maximum  shear  at 

,        wL 
the  ends  =  -=— 

A  crane  travels  on  a  girder  with  two  wheels  equally  loaded  and 

separated  by  a  constant  dis- 
tance. The  maximum  bending 
moment  is  under  one  wheel 
when  this  wheel  is  as  far  from 
.i r #  one  SUpp0rt  as  the  center  of 

Fig.  30  —  Two  Equal  Rolling  Loads  at    gravity   of    the   total   load   is 
Fixed  Distance  Apart  from    ^     opposite     gupp()rt 

Referring  to  Fig.  30,  if  a  is  less  than  0.586  L, 
M  = 


EXTERNAL  FORCES  37 

The  maximum  moment  will  occur  twice  on  the  span  as  the  load 
moves  along,  the  moments  being  equal  and  distant  one-fourth  a 
from  the  middle  of  .the  span.    The  maximum  shear  is  at  one  end 
when,  one    of    the   wheels  is 
directly  over  the  edge  of  the 
support.  j  (4) 

In  Fig.  31  is  illustrated  the               — fe- 
case  of  two  unequal  loads  at       ff -* * —  L -~" If 

fixed  distance    apart,  moving    „.    . 

3    Fig.  31  —  Two  Unequal  Rolling  Loads 
across  a  span.     Ihe  maximum  at  flxe(^  Distance  Apart 

moment  is  under  the  heavier 

load.    The  maximum  end  shear  is  under  the  heavier  load  when  it 

is  over  the  edge  of  the  support. 

Let        w  =  weight  of  lighter  wheel  load  (A). 
W  =  total  load  =  A  +  B. 
a  =  distance  center  to  center  of  wheels. 
y  =  distance  from  heavier  wheel  to  mid-span. 
M max  =  maximum  bending  moment  under  heavier  load. 


JlSr- 


In  Fig.  32  is  shown  a  graphical  method  for  ascertaining  the 
bending  moment  when  a  load  occupies  a  definite  length  on  the 
beam.  The  triangle  is  first  drawn  as  though  the  entire  weight 
was  concentrated  at  the  center  of  gravity  of  the  load.  Drop  ver- 
tical lines  from  the  ends  of  the  load  to  intersect  the  triangle. 
Connect  the  points  of  intersection  by  a  straight  line.  Use  this 
line  as  the  base  of  a  parabola,  which  is  then  constructed  as  shown. 
This  method  is  also  used  if  one  end  of  the  load  rests  on  one 
abutment. 

Overhanging  Beams 

When  beams  overhang  one  or  two  supports  the  methods  for 
obtaining  reactions,  bending  moments,  and  shears  are  no  different 
from  those  used  for  cantilever  beams  and  beams  resting  freely 
on  two  supports.  The  three  examples  following  are  from  Greene's 
"Structural  Mechanics"  (3d  ed.). 


38 


PRACTICAL  STRUCTURAL  DESIGN 


Fig.  32  —  Graphical  Method  for  Concentrated  Load  with  Wide  Bearing 

Studying  Fig.  33  we  see  that  the  reactions  are  as  follows: 
750  x  25 


20 


937.5  Ibs. 


R*  =  750- 937.5  ^  -187.5  Ibs. 

The  weight  of  the  beam  has  been  neglected  and  the  load  at  the 
extreme  left  tends  to  revolve  the  beam  around  Ri  so  an  additional 


20'- 


R2—IB7.S 


Fig.  33  —  Concentrated  Load  on  Short  Overhanging  Beam 

load  of  187.5  Ibs.  will  be  required  at  #2  to  hold  down  the  right 
end.    The  negative  sign  indicates  an  upward  pull. 


EXTERNAL  FORCES  39 

The  bending  moment  at  Ri  is 

M  =  -750  x  5  =  -3750  ft.  Ibs., 

indicating  a  negative  bending  moment,  tending  to  make  the 
beam  convex  at  this  support,  showing  tension  to  exist  in  the 
upper  part  and  compression  in  the  lower  part  of  the  beam.  At 
section  x-xi 

M  =  -#2  X  10  =  -187.5  x  10  =  -1875  ft.  Ibs. 
also  negative  because  R2  is  negative. 

Check  the  moment  at  Ri  as  follows,  using  R2: 
M  =  -187.5  x  20  =  -3750  ft.  Ibs. 

This  beam  has  a  negative  bending  moment  at  all  points,  which 
indicates  tension  above,  and  compression  below,  the  neutral  axis. 

«  R2— eoo  itx 


L-  ........................  20'- 


J...V.^....J 
fi,"750/bs. 

Fig.  34  —  Concentrated  Load  on  Long  Overhanging  Beam  with  Supports 
Close  Together 

A  positive  bending  moment  indicates  tension  below,  and  compres- 
sion above,  the  neutral  axis. 
In  Fig.  34  the  reactions  are 

ft_  150X25  =  750,bs 

o 
fl2  =  150  -  750  =  -600  Ibs. 

Note  that  the  divisor  is,  in  all  cases,  the  distance  between  sup- 
ports. Both  reactions  are  large  compared  with  the  load,  showing 
the  absurdity  of  considering  a  beam  to  be  fastened  at  supports 
when  it  runs  only  a  short  distance  into  a  wall.  No  beam  should 
be  considered  as  tied  unless  it  runs  into  a  wall  a  couple  of  feet  at 
least  and  actually  carries  enough  load  to  counteract  the  amount 
of  negative  reaction. 

Several  loads  are  shown  in  Fig.  35  on  a  beam  resting  on  two 
supports  and  overhanging  one  of  the  supports. 

D      100x18+200x16+150x13+300x11+50x8+80 


16 
880  -  665.625  =  214.375  Ibs. 


.__  A0_  „ 
=  665.62D  Ibs. 


40  PRACTICAL  STRUCTURAL  DESIGN 

Fig.  36  illustrates  an  example  given  in  Volume  1  of  "  Building 
Construction,"  edited  by  F.  M.  Simpson,  in  which  the  weight  of 
the  beam  is  considered.  The  total  length  of  the  beam  is  28  ft.  and 
the  weight  is  100  Ibs.  per  lineal  foot. 


Fig.  35  —  Beam  Carrying  Several  Concentrated  Loads  with 
Short  Overhang  at  One  End 

p      23  X  1000  +  14  x  1200  +  9  X  2800  -  5  x  500 

ti\  =  —  20  —  ~" 

Rz  =  5500  -  3125  =  1875  Ibs. 

For  all  the  overhanging  beam  cases  the  amount  and  location 
of  bending  moment  and  shear  at  any  point  on  the  beam  may  be 
found  by  the  rules  previously  given  for  cantilever  beams  and 
beams  resting  on  two  supports.  Notice  that  the  distance  from 
either  reaction  to  the  center  of  gravity  of  the  weight  of  the  beam 
is  equal  to  the  distance  from  the  other  reaction  to  the  center  of 
gravity  of  the  weight  of  the  beam  for  uniformly  distributed  loads 
covering  the  whole  beam.  This  applies  as  well  to  the  team  alone, 
for  the  weight  of  a  beam  is  uniformly  distributed. 

In  Fig.  37  the  beam  weighs  20  Ibs.  per  lineal  foot  =  21  x  20  =  420 
Ibs.  Half  is  carried  on  each  support,  for  the  overhang  is  equal  at 


19'- 


.-4.-  ........  -  ...................  -20'.  ..............  _  ............  L  ......  5'  .....  j 

R2-2375/b& 


Fig.  36  —  Beam  Carrying  Several  Concentrated  Loads,  with 
Both  Ends  Overhanging 

either  end.    Each  support  also  carries  the  concentrated  load  near- 
est to  it  in  this  particular  example. 

Rl  =  R2  =  210  +  250  =  460  Ibs. 

On  the  moment  diagram  the  upper  curved  line  (parabola  with 
vertex  at  the  support)  under  the  overhanging  end  shows  moment 


EXTERNAL  FORCES 


41 


due  to  the  cantilever  end  of  the  beam.  The  straight  line  under 
it  shows  moment  due  to  the  concentrated  load  on  the  extreme 
end.  The  lower  slightly  curved  lines,  AC  and  BD,  represent  the 
combined  moments  under  the  cantilever  ends.  This  line  at  each 
point  is  the  sum  of  the  two  moments,  so  is  a  mean  between  the 
parabola  and  straight  line. 

The  bending  moment  at  the  middle  of  the  span  between  the 
two  supports  is  found  as  follows,  there  being  a  positive  and  a  nega- 
tive moment  to  consider: 

-M  =  250  x  10.5  +  (10.5  x  20)  x  5.25  =  -3827.5  ft.  Ibs. 
+M  =  460  X  7.5  =  3450  ft.  Ibs. 
Actual 

M  =  +M  -  M  =  -3827.5  +  3450  =  -377.5  ft.  Ibs. 

This  negative  mo- 
ment is  set  off  at  the 
middle  of  the  span 
measuring  down  from 
the  line  A  B.  The  pa- 
rabola CED  is  drawn. 
The  bending  moment 
at  any  point  is  found 
by  scaling  the  length 
intercepted  between 
the  line  AB  and  the 
bounding  line  ACEDB 
of  the  bending  moment 
diagram.  All  lengths 
measured  horizontally 
are  distances  and  all 
lengths  measured  ver- 
tically are  forces. 


t— 1 


-T-  I—L-.V- J 

U^-~~ l5'~*--~ — ipt 


Shear  Diagram 


Fig.  37  —  Graphical  Method  for  Beam  with  Two 
Overhanging  Ends 


When  the  positive  moment  is  greater  than  the  negative  moment 
the  point  E  is  set  off  above  the  line  AB,  so  the  parabola  in  such 
case  is  partly  above  and  partly  below  this  line.  The  curve  above 
indicates  positive  and  the  curve  below  indicates  negative  moment. 
The  maximum  moment  is  where  the  shear  changes  sign.  Where 
the  moment  curve  crosses  the  line  AB  there  is  no  moment,  this 
point  being  termed  the  "  point  of  reverse  moment "  or  "  point  of 
contraflexure,"  or  "  point  of  inflection." 


42  PRACTICAL  STRUCTURAL  DESIGN 

The  shear  diagram  is  evident.  After  drawing  all  the  diagrams 
the  student  should  check  the  moment  at  different  sections  by  means 
of  the  shear  diagram.  Drop  a  vertical  line  through  the  shear 
diagram  from  any  point  on  the  beam.  The  area  of  the  shear  dia- 
gram between  this  section  and  the  nearest  support  is  equal  to  the 
bending  moment  at  the  section.  When  a  beam  rests  on  two  sup- 
ports and  the  moment  is  desired  at  any  point  the  beam  is  assumed 
to  be  fixed  there  and  to  be  pushed  up  by  a  force  equal  in  amount 
to  the  reaction.  Thus  it  is  a  cantilever  beam,  and  by  reference  to 
Fig.  3  it  will  be  found  that  for  a  cantilever  beam  the  moment 
at  any  section  is  equal  to  the  area  of  the  shear  diagram  between 
that  section  and  the  free  end. 

The  principle  of  the  lever  applies  in  all  cases  and  moment  effects 
are  additive;  therefore  the  effect  of  additional  concentrated  loads 
on  the  beam  may  be  readily  found.  This  is  recommended  as  an 
exercise,  arithmetical  computations  being  checked  by  graphical 
methods. 

In  the  figures  an  arrow  point  indicates  the  center  of  gravity 
of  the  bearing  area.  The  clear  span  is  S  and  the  length  of  the 
beam  is  L\.  The  moment  span  is  L.  To  simplify  all  computa- 

T 

tions  use  L 

, , 

and  instead  of  M 


use  the  formula  M  = 


2 

w  X  S  X  I/i 

~8~~ 
wL2 


the  average  length  being  used  in  all  cases,  as  it  is  close  enough 
for  all  practical  purposes. 

In  examples  involving  loads  concentrated  at  some  point  one 
side  of  the  middle  of  a  span  the  distance  to  the  nearer  support  has 
been  termed  a  and  the  distance  to  the  farther  support  b.     Then 
Pab 

M  =  ~T- 

The  custom  in  modern  text  books  is  to  use  the  letter  a  for  the 
shorter  length  and  designate  the  longer  length  by  describing  it 
as  the  difference  between  L  and  a.  Thus 

u      Pa(L  ~  a) 
M  -      -- 


EXTERNAL  FORCES  43 

The  older  method  was  given  for  the  reason  that  it  is  so  frequently 
met  with  in  handbooks  and  trade  papers,  but  the  method  of  modern 
text  books  is  preferable  and  should  be  used  by  the  student  in  his 
work. 

Equivalent  Distributed  Loads 

A  convenient  method  to  use  in  figuring  bending  moments  when 
one  or  more  concentrated  loads  must  be  considered  in  addition 
to  a  uniformly  distributed  load,  is  to  reduce  the  concentrated 
loads  to  equivalent  distributed  loads.  Suppose  we  take  the  ex- 
pression last  given  for  the  effect  of  a  concentrated  load  at  some 
point  of  the  beam: 


The  problem  is  to  find  the  value  of  W,  the  uniformly  distributed 
load. 

W  T 

Arrange  it  thus,  M  =  —5— 

o 

Eliminating  fractions,      8M  =  WL 

Dividing,  W  =  -^— 

LI 

The  student  can  see  that  after  obtaining  the  bending  moment 
for  the  concentrated  load  the  bending  moment  had  only  to  be 
equated  to  that  for  a  uniformly  distributed  load.  If  he  does  a 
little  thinking  he  will  see  that  if  the  concentrated  load  is  off  center 
very  far  the  bending  moment  is  greater  than  it  is  at  the  center  of 
the  span,  yet  the  equation  of  the  uniformly  distributed  load  was 
made  on  the  basis  of  the  maximum  moment  being  at  the  center  of 
the  span. 

The  method  of  equivalent  uniformly  distributed  loads  is  in 
common  use  in  many  designing  offices,  but  only  because  it  saves 
a  little  time  and  because  beams  come  in  stock  sizes.  It  always 
gives  a  trifle  larger  beam  than  is  necessary,  so  it  is  a  safe  method 
to  use.  When  the  greatest  possible  economy  is  desired,  or  the 
beam  size  selected  is  on  the  border  line  between  a  heavy  and  a 
light  beam,  the  exact  method  should  be  used  to  obtain  the  size, 
as  thereby  considerable  saving  may  be  effected.  The  exact  method 
should  be  used  also  when  a  built-up  girder  is  to  be  designed. 

The  uniform  load  has  been  found.  Divide  it  by  the  concentrated 
load  and  get  a  factor  we  will  call  m.  Then  divide  the  span  by  the 


44  PRACTICAL  STRUCTURAL  DESIGN 

distance  from  the  concentrated  load  to  the  nearest  support  and 
call  the  result  x. 

W  L 


Example.  —  A  beam  weighing  200  Ibs.  on  a  span  of  12  ft.  carries 
a  concentrated  load  of  750  Ibs.  3  ft.  from  one  end.  What  is  the 
equivalent  distributed  load?  What  is  the  bending  moment? 

,  ,      Pa(L  -a)      750  x  3  x  9        ___  _  ,,   lu 
Ans.        M  =  —  *-=  -  -  =  --  77;  --  =  1687.5  ft.  Ibs. 
L  12 

8M8x  1687.5 


W      1125 
m  -  ~P  =  750-  =  L5' 

L      12 

x  =  —  =  —  =  4. 
a       3 

The  equivalent  distributed  load  producing  a  bending  moment 
equivalent  to  the  bending  moment  produced  by  the  concentrated 
load  =  1125  Ibs.  and  to  this  must  be  added  the  weight  of  the 
beam,  200  Ibs.  The  total  bending  moment  is 

WL      1325X2 


The  student  is  advised  to  compute  a  table,  following  the  above 
example,  with  the  concentrated  load  assumed  to  be  placed  at  vari- 
ous points  on  the  beam,  the  table  giving  values  of  m  and  x,  to  be 
used  in  shortening  labor  in  future  work. 

Such  tables  are  in  use  giving  values  of  m  and  x  for  a  dozen  or 
more  points  on  a  beam.    The  following  table  gives  these  values 
for  ten  points: 
When  x  =  2  m  =  2 

x  =  3  m  =  1.78 

x  =  4  m  =  1.5 

x  =  5  m  =  1.28 

x  =  6  m  =  1.11 

x  =  7  m  =  0.98 

x  =  8  m  =  0.875 

x  =  9  m  =  0.79 

x  =  10  m  =  0.72 


EXTERNAL  FORCES 


45 


In  using  the  table  first  find  the  value  of  x  by  dividing  the  total 
span  by  the  distance  from  the  nearest  support  to  the  load.  Then 
multiply  the  load  by  m  in  the  table  opposite  the  value  found  for 
x.  Do  this  for  each  concentrated  load  in  turn,  add  the  weight  of 


Loading 

Max/mum 
Bending 
Momenr 

Relative 
Strength 

Relative 
Deflection 
in  terms  of 
stress 

1                         ® 

i 

/ 

8 

i 

1               PV            1 

M=WL 

4 

2-4 

$          / 

2 

<p~~~        L          n 

© 

j-4--*l~--M 

^=a 

2 

0-8 

1              W             \ 

A,/    H//_ 

M=5 

1     (p)     p 

iirtij 

*** 

i 

0.4 

^           ^ 

1             M^             I 

M=  — 

3 

0.2 

|  L-  | 

Fig.  38  —  Reference  Table  Showing  the  Strength  and 
Stiffness  of  Beams 

the  beam,  and  then  find  the  bending  moment  for  the  total  uniformly 
distributed  load  so  obtained. 

The  method  of  equivalent  uniformly  distributed  loads  is 
applicable  only  to  bending  moments.  When  a  load  is  uniformly 
distributed  the  end  reactions  are  equal  and  the  shear  is  always 
equal  to  the  reaction,  so  if  the  reactions  and  shears  are  considered 
at  each  end  as  being  one-half  the  equivalent  distributed  load  the 
beam  may  be  weak  in  shearing  resistance  and  the  supports  may  be 
improperly  designed.  It  is  necessary,  therefore,  to  compute  the 


46  PRACTICAL  STRUCTURAL  DESIGN 

reactions  by  the  exact  method  and  at  the  same  time  find  the 
maximum  shear.  The  beam  will  be  designed  for  maximum  shear 
and  the  supports  will  be  properly  taken  care  of. 

The  accompanying  table  of  strength  and  stiffness  of  beam  is 
valuable  for  daily  reference  in  beam  calculations.  The  subject 
of  deflection  will  be  taken  up  later.  This  table,  Fig.  38,  takes 
as  a  basis  the  uniformly  distributed  load  on  a  beam  resting  freely 
on  two  supports.  In  the  first  column  is  shown  the  loading  con- 
dition; in  the  second  column  the  formulas  for  ascertaining  the 
bending  moments;  in  the  third  column  the  relative  loads,  and  in 
the  fourth  column  the  relative  deflection  due  to  these  loads.  For 
example,  the  cantilever  beam  carrying  a  concentrated  load  at  one 
end  will  support  only  one-eighth  the  load  the  same  size  beam  with 
the  same  span  will  carry  when  freely  supported  at  the  two  ends. 
The  deflection  under  this  load  will  be  3.2  times  the  deflection  of 
the  freely  supported  beam  carrying  8  times  the  load.  The  uni- 
formly distributed  load  on  a  beam  securely  fastened  over  supports 
is  1.5(3/2)  times  the  load  carried  on  the  same  beam  on  the 
same  span  when  freely  supported  and  the  deflection  is  greatly 
lessened,  being  only  0.3  the  deflection  of  the  freely  supported 
beam  carrying  two-thirds  the  load  of  the  restrained  beam. 

Restrained  Beams 

In  Fig.  39  is  shown  a  beam  tied  into  the  supports.  This  is  known 
as  a  restrained  beam.  A  restrained  beam  carrying  one  centrally 
concentrated  load  will  be  first  considered. 


•Q- 


Fig.  39 — Beam  Resting  on  Fig.  40 —  Beam  Deflected  under 

Supports  Loads 

When  a  beam  is  simply  supported,  that  is  rests  on  supports 
without  being  fastened  in  place,  it  deflects  under  load  as  shown 
in  an  exaggerated  manner  in  Fig.  40,  so  the  ends  AB  and  CD 
slope  and  are  no  longer  vertical.  At  E  there  is  compression  and 
at  F  there  is  tension,  but  no  tensile  or  compressive  stresses  exist 
at  A,  B,  C  and  D. 

When  a  beam  having  a  uniform  moment  of  inertia  (that  is,  a 


EXTERNAL  FORCES 


47 


y 

I 


Fig.  41 


beam  symmetrical  in  form  with  material  uniform  throughout) 

is  restrained,  the  ends  have  no  slope  when  the  beam  carries  a  load. 

The  case  is  shown  in  Fig.  41,  where  the  ends  are  extended  to 

some    distance,    b,    where    a 

load  is  placed  which  has  sufn-     I 

cient  weight  to  hold  the  ends  Q 

in  the  original  positions.     '- 

Tension    under    such    condi-     L 

tions  exists  at  A  and  C  with 

compression  at  B  and  D.    At 

the   point   where    the   beam 

ceases  to  be  horizontal  and  bends  down  there  is  neither  tension 

nor  compression,  the  only  existing  force  being  shear.    This  point 

is  termed  the  "  point  of  contraflexure,"  the  "  point   of  reverse 

moment,"  or  the  "  point  of  inflection." 

In  Fig.  42  the  shaded  tri- 
angles represent  the  mo- 
ments of  the  actual  center 
load  and  the  two  assumed 
end  loads.  These  loads,  as 
well  as  the  length  6,  may 
actually  exist,  but  the  same 
effect  will  be  obtained  by  riveting  or  otherwise  fastening  the  ends 
of  the  beams  to,  or  in,  solid  supports;  therefore  the  loads  and  the 
moment  areas  beyond  the  point  of  support  are  said  to  be  imagi- 
nary or  assumed.  The  condition  created  is  that  of  a  simply  sup- 
ported beam,  having  a  length  measured  between  the  points  of 
contraflexure,  carried  on  the  ends  of 
two  short  cantilevers.  An  expression 
must  be  found  for  the  force  creating 
such  a  condition  and  also  for  the 
lengths  of  the  cantilevers,  that  is,  the 
distance  from  the  point  of  support  to 
the  point  of  contraflexure. 

In  Fig.  43  let  the  triangle  ABC  re- 
present the  moment  area  due  to  a 
concentrated  load  at  midspan  of  a 
freely  supported  beam,  AC.  The  two  end  triangles  AGF  and  CDE 
are  the  moment  areas  of  the  loads  causing  the  restraint.  An  in- 
spection will  show  that  the  combined  areas  of  the  two  end  triangles 


Fig.  42 


48  PRACTICAL  STRUCTURAL  DESIGN 

must  equal  the  area  of  the  triangle  FEE  and  that  the  points  of 
contraflexure  lie  vertically  beneath  E  and  F. 

Assume  a  rectangle  AGDC  placed  on  the  span  AC.  The  area 
of  this  rectangle  times  the  distance  to  the  center  of  gravity  from 
either  support  must  equal  the  area  of  the  triangle  ABC  times  the 
distance  of  its  center  of  gravity  from  the  same  support.  This  is 
known  as  equating  moment  areas,  or, 

(AC  x  AG)  x  y  =  — ^—  x  y. 

Let  AG  =  x.  The  span,  AC,  is  common  to  both  sides  of  the 
equation  and  the  length,  y,  of  the  moment  arm  is  one-half  of  AC, 
so  may  be  eliminated,  as  it  also  is  common  *to  both  sides  of  the 
equation,  which  is  treated  as  follows; 

AC  x  Z 

Eliminating  y,  AC  x  x  =  — = 

^ 

Eliminating  the  common  factor  AC,  x  =  — 

£i 

V 

From  the  similarity  of  triangles,  since  AG  =  DC  =  x  =  -& 

then  AF  =  FB,  and  CE  =  EB,  for  F  lies  in  the  line  AB  and  E 
lies  in  the  line  CB,  and  the  line  GD,  parallel  to  AC,  intersects  the 
line  AB  in  F  and  the  line  EC  in  E, 

The  length  EF  =  GF  +  ED;  therefore. the  area  of  the  triangle 
FBE  =  area  of  triangle  AGF  +  area  of  triangle  CDS.  The  length 

AC 

GF  =  ED  =  -^— ;    therefore  the   points  of  contraflexure  are   \L 

from  the  supports. 

Since   Z  =  —^-  and  x  =  •=> 

PT  PT 

then        x  =  negative  moment  =  —  \  x  —r-  =  — ~— 

The  positive  moment 

Z      I      PL         PL 

=2=2X  4  =+-r 

PI        PT  PT  7 

Let    ^  =  ^P;  then  PI  =  ^  and  I  =  ^  (eliminating  P). 
4  o  ^  ^ 

Since       I  =  -^  and  I  +  2a  =  L, 
L  -I      L 


EXTERNAL  FORCES  49 

The  uniformly  loaded  beam  restrained  at  the  ends  is  shown  in 
Fig.  44.  The  reasoning  follows  that  for  the  beam  carrying  one 
center  concentrated  load.  The  bending  moment  diagram,  however, 

WL 

is  a  parabola  with  height,  Z  =  — — >  the  area  of  which  is  equal  to 

o 

the  product  of  the  base  times  two- 
thirds  the  height ;  then 


Eliminate  like  quantities  from  the 
two  sides  of  the  equation  and 
2Z 

*-T 

Then  x  =  negative  moment  = 

2      WL         WL 
~  3  x  IT  =  ~  l2~'  at  ea     e 

The  center  positive  moment  =  «  x  — ^—  =  +  -^j— 
o          o  2A 

Now  take  unit  load  per  lineal  foot  =  w. 

"A. '''.        WJ2        WL2      t  WL2          .   ,„         L2  ,  ,.      .        ..  x 

Let     ~^~  =  ~^r>  tnen  wl  =  ~5~  an(i '  =  "5~  (eliminating  «?). 

o  Z4  o  o 

z      /?  =  A      L 

\  3       V3      1-732 
When  L  =  1,1  =  -^  =  0.5773L. 

1. 1  oZ 

Since    Z  =  0.5773L  and  Z  +  2a  =  L, 
a  =  L_^  1.0000- 0.5773 

Continuous  Beams 

Continuous  beams,  that  is  beams  running  over  a  number  of 
supports,  are  designed  by  methods  which  are  an  extension  of  the 
principles  used  for  overhanging  beams  and  restrained  beams. 
The  only  instance  of  continuous  beams  in  wood  construction  ap- 
pears in  the  placing  of  floors  over  joists  or  closely  spaced  beams. 
On  account  of  the  excessive  deflection  of  wood  this  is  not  justifi- 
able, for  the  theory  underlying  continuous  and  restrained  beams 


50  PRACTICAL  STRUCTURAL  DESIGN 

requires  that  the  supports  be  immovable.  The  slightest  settle- 
ment causes  increased  stress  and  sometimes  a  reversal  of  stress. 
As  it  is  merely  a  matter  of  properly  designed  connections,  con- 
tinuous girders  and  beams  are  sometimes  used  in  steel  structures. 
They  are  not  common,  however. 

The  maximum  stresses  are  invariably  over  the  supports,  and 
lack  of  realization  of  this  fact  has  caused  distress  to  some  de- 
signers. The  principle  of  continuous  beams  finds  application  in 
reinforced  concrete  work.  Owing  to  the  monolithic  character  of 
reinforced  concrete  there  is  no  other  proper  way  for  designing  in 
this  material.  To  assume  that  the  bending  moment  on  a  span 

=  —  —  does  not  make  it  so,  and  designers  who  assumed  that  the 

o 

greater  stiffness  thus  secured  would  permit  the  use  of  a  smaller 
moment  over  supports  suffered  in  reputation  thereby. 

By  methods  involving  the  use  of  higher  mathematics  it  can  be 
proved  that  the  sum  of  the  moment  in  the  middle  of  the  span  and 

the  moment  at  one  support  =  -5—    In  the  study  of  restrained 

o 

beams  this  has  been  shown,  for,  disregarding  the  signs, 

WL      WL  _  3TFL  _  WL 

12   +   24    =:     24  8  ' 

The  smaller  moment,  however,  is  in  the  span  and  the  larger 

WL 
moment  is  over  the  support.     To  assume  M  =  —^  does  n°t  make 

the  moment  over  the  support  =  -^r-  based  on  the  following  reason- 

WL      WL     3TFL      2WL      WL 
ing;  — j^-  =  -g| ^4"  =  -^"     The  moment  over  the 

WL 
support  is  -vo~'  or  nearly  this  amount,  no  matter  what  assumption 

may  be  made  for  the  bending  coefficient  in  the  middle  of  the 
span. 

When  the  moment  of  inertia  of  the  beam,  or  slab,  is  constant, 

the  tension  over  the  support  is  that  due  to  a  moment  =  — ^r-> 

if  the  span  is  designed  for  M  =  +  -jo~'  However,  if  the  amount  of 
steel  is  reduced  so  the  resisting  moment  will  be  barely  sufficient 


EXTERNAL  FORCES 


51 


45  —  Coefficients   for   Maximum    Negative 

Bending  Moments  for  Uniform  Loads  over 

the  Supports  of  Continuous  Beams  with 

Equal  Spans.     M  =  CwP,  in  which 

C  =  coefficient. 

Example:  M  =  ~  =  0.125  wP. 


to  take  care  of  this  amount  of  bending  moment,  the  moment  of 
inertia  is  thereby  altered.  Changing  the  moment  of  inertia  has 
the  effect  of  greatly  increasing  the  tension  in  the  steel  over  the 
supports.  The  posi- 
tive and  negative 
moments  should  total 
%WL  and  not  \WL, 
this  in  effect  making 
the  positive  and  nega- 
tive moments  for  in- 
terior spans  and  sup- 
ports =  iW£- 

When  one  panel  is 
loaded  and  an   adja- 
cent  panel  is  not 
loaded  there    will   be 
an  uplift  in   the   un- 
loaded panel,  for  it  opposes  only  the  dead  load  to  the  combined 
dead  and  live  load  on  the  loaded  panel.     To  make  the  positive 
and  negative  moment  coefficients  in  each  panel  equal  gives  the 
necessary  stiffness  and  increased  weight. 

Assuming  spans  equal  in  length  and  loaded  uniformly,   the 

negative  bending  moment 
coefficients    to    use    over 
supports  are  shown  in  Fig. 
45.     Each    square    repre- 
sents a  support   and  the 
coefficients    are    given    as 
decimal  instead    of    com- 
mon fractions.     The  mo- 
Fig.  46  —  Coefficients  for  Maximum  Positive  ,    ,  i 
Bending  Moments  for  Uniform  Loads  on       ments  are  constant  at  the 
Equal  Spans   of   Continuous  Beams.          edges  and  across  the  tops. 
M  =  Cwl-,  in  which  C  =  coefficient.              Assuming    spans   equal 
Examples:  M  =  ^  =  0.125  wP.            in  leri&h  and  loaded  uni- 
formly, the  positive  bend- 
ing moment  coefficients  to  use  for  the  spans  are  shown  in  Fig.  46. 
The  coefficients,  however,  are  the  theoretical  coefficients  for  beams 
with  constant  moment  of  inertia.    They  should  not  be  used  for 
the  reasons  given  above,  for  it  is  best  to  have  the  positive  and 
negative  moments  equal. 


52  PRACTICAL  STRUCTURAL  DESIGN 

Assuming  spans  equal  in  length  and  loaded  uniformly,  the 
coefficients  to  use  in  figuring  the  reactions  on  the  supports  are 
given  in  Fig.  47.  These  coefficients  give  the  total  reaction  due  to 
the  load  from  the  middle  of  one  span  to  the  middle  of  the  adjacent 

span,   one    full   panel 
length.    The  moments 
on  each   side    of  the 
support   differ,   as 
shown  in  Fig.  46,  so 
the  shear  at  the  edge 
of  the  supports  is  pro- 
portional  to  the  mo- 
Fig.  47  —  Coefficients  for  Reactions  for  Beams   ment     coefficients    in 
under  Uniform  Load  over  Several  Equal  Spans       fup    annn<s        TT«nallv 
and  Freely  Supported  at  the  Ends,     fl  =  Cwl.      ™*    SpanS/    .    JsUfly' 

however,  it  is  safe  to 

use  half  the  reaction  for  the  shear  on  each  side  of  a  support.  The 
three  figures  are  all  based  on  the  assumption  that  the  ends  of  the 
beams  are  freely  supported. 

Spans  are  not  always  equal  in  length  for  continuous  beams 
and  the  beams  are  not  always  uniformly  loaded.  A  complete 
discussion  of  such  conditions  is  best  treated  graphically  and  will 
be  taken  up  in  another  chapter. 

When  spans  are  unequal  the  reactions  for  continuous  beams 
must  be  computed  for  each  span  separately.  The  total  reaction 
on  any  intermediate  support  is  equal  to  the  sum  of  the  reactions 
for  the  adjacent  spans. 

Let  MI  =  moment  at  left  end  of  span. 
Mz  =  moment  at  right  end  of  span. 
Ri  =  reaction  at  left  end. 
Rz  =  reaction  at  right  end. 
w  =  load  per  lineal  foot. 

Z  =  span.    When  the  moment  is  in  foot  pounds  the  span 
is  in  feet.  /2 

R,  x  I  -  M2  =  -Mi  +  ^> 

by  taking  moments  about  Rz  from  which 

wl      Mj  -  M,        , 
Rl  =  ~2 1 

„       wl      M2-M! 


EXTERNAL  FORCES  53 

The  reaction  from  a  continuous  slab  is  carried  by  the  beam  and 
constitutes  the  uniform  load  on  the  beam,  plus  the  weight  of  the 
beam.  This  in  turn  is  carried  by  the  girder  and,  together  with 
the  weight  of  the  girder,  constitutes  the  uniform  load  on  the  girder. 
The  girder,  if  continuous,  transmits  to  the  columns  the  effect  of 
the  unbalanced  moments  over  supports,  of  slab,  beam,  and  girder. 
Columns  must  be  proportioned  to  carry  this  load  made  up  of  the 
direct  weights  plus  the  unbalanced  moments  in  the  system  of 
framing. 

Building  ordinances  require  that  the  effects  of  unbalanced 
moments  must  be  considered  in  design.  The  author  in  his  own 
practice  designs  in  this  manner.  It  is  the  practice  of  all  reputable 
designers.  The  majority  of  reinforced  concrete  buildings  in  the 
past  have  been  designed  by  firms  engaged  in  the  business  of  sell- 
ing steel.  A  large  part  of  the  work  of  the  author  for  several  years 
consisted  in  checking  these  "  free  "  (so-called)  designs  and  he  has 
not  found  one  in  which  anything  more  than  the  direct  load  was 
considered.  The  practice  being  somewhat  common  and  struc- 
tural engineers  in  building  departments  having  passed  designs  so 
prepared,  he  has  had  no  other  alternative,  for  it  was  what  is  termed 
"  common  practice."  After  the  date  on  which  the  present  book 
is  placed  on  the  market  he  will  approve  no  more  designs  following 
the  old  "  common  practice  "  and  will  force  designers  to  fully 
recognize  the  effect  of  unbalanced  moments. 

The  "  free  design  "  is  usually  given  in  the  following  way.  An 
architect  is  employed  by  an  owner  to  prepare  plans  for  a  rein- 
forced concrete  structure.  The  architect  not  being  an  engineer 
specialist  has  a  choice  of  either  employing  a  reputable  engineer 
to  prepare  the  engineering  design,  for  which  he  must  pay  a  con- 
siderable part  of  his  own  fee  if  he  has  no  engineer  employed  on 
salary,  or  he  merely  prepares  general  drawings  and  in  his  speci- 
fications states  that  competitive  engineering  plans  will  be  accepted. 
Sometimes  he  fixes  the  stresses  and  other  necessary  designing  speci- 
fications, but  more  commonly  omits  to  do  so.  Many  architects 
believe  that  reinforced  concrete  design  is  as  standard  as  steel 
design,  which  is  not  the  case. 

Contractors  who  bid  on  the  work  apply  to  steel  selling  con- 
cerns for  designs.  These  are  prepared  and  the  contractor  is  given 
a  lump  sum  price  for  the  steel,  which  includes  the  cost  of  making 
the  design.  Frequently  the  amount  of  concrete  is  guaranteed  by 


54  PRACTICAL  STRUCTURAL  DESIGN 

the  steel  salesman.  Sometimes  the  above  procedure  is  varied  by 
a  draftsman  being  employed  to  make  the  building  plans  and  the 
owner  takes  up  the  design  of  the  structure  directly  with  some  steel 
salesman.  In  this  case  the  plans  are  furnished  without  cost  to 
the  owner,  provided  the  firm  supplying  the  plans  sells  the  steel. 
A  price  is  fixed  for  the  plans  in  case  the  owner  purchases  the  steel 
elsewhere.  The  price  for  the  design  must  then  be  added  to  bids 
for  the  steel,  which  usually  results  in  the  owner  purchasing  the 
steel  from  the  firm  giving  him  the  "  free  "  design.  There  is  severe 
competition  in  the  steel-selling  game,  the  result  being  that  few 
designs  thus  made  are  as  good  as  they  should  be. 

The  designers  employed  by  steel  salesmen  are  first-class  men 
in  nearly  every  instance,  but  in  order  to  hold  their  positions  they 
work  for  their  employers  rather  than  for  the  owners,  which  is  quite 
natural.  Few  of  these  men,  when  they  finally  go  into  business  for 
themselves,  design  exactly  as  they  were  forced  to  design  when  they 
were  required  to  assist  in  selling  steel.  If  the  owner  employs  a 
reputable  engineer  in  private  practice  to  do  the  engineering  work 
and  pays  him  a  fee  in  addition  to  that  paid  to  the  architect,  he 
will  receive  designs  which  conform  to  the  best  modern  practice  and 
all  contractors  will  bid  on  these  designs.  The  difference  in  cost 
between  the  certainty  of  good  work  and  the  uncertainty  affect- 
ing the  "  skinning  "  of  designs  by  steel  salesmen  will  be  very  small. 
Frequently  the  competition  between  contractors  will  entirely 
wipe  out  the  difference. 

Owing  .to  the  large  number  of  unsatisfactory  designs  received 
under  the  competitive  method  of  having  steel  salesmen  furnish 
the  engineering  work,  and  to  the  failures  occurring  during  con- 
struction, the  practice  should  be  abandoned.  The  average  owner, 
and  many  architects,  consider  the  engineer  as  an  expert  juggler 
with  figures  and  a  sort  of  human  attachment  to  a  slide  rule  and 
table  book.  When  a  steel-selling  concern  boasts  of  the  ability  of 
the  engineers  employed  by  it  the  engineers  are  really  only  highly 
trained  men  doing  clerical  work,  for  computing  the  strength  of 
parts  is  nothing  more.  The  engineer  really  is,  or  should  be  so 
considered,  a  man  whose  only  interest  is  the  safeguarding  of  the 
interests  of  his  employer.  No  man  can  serve  two  masters,  and  the 
designer  who  is  trying  to  help  his  employer  make  the  largest  pos- 
sible profit  on  the  sale  of  steel  cannot  be  a  disinterested  designer 
for  the  owner  who  does  not  employ  him  directly  but  merely 


EXTERNAL  FORCES        .  55 

obtains  his  technical  services  through  a  salesman  as  an  inter- 
mediary. 

Sometimes  architects  and  owners  obtain  competitive  designs 
on  specifications  prepared  by  a  reputable  engineer  in  private 
practice  and  insert  in  the  specifications  a  clause  that  the  success- 
ful bidder  must  deposit  some  definite  amount  to  pay  the  cost  of 
having  the  designs  checked  by  the  engineer,  whose  name  is  given. 
The  author  does  a  great  deal  of  work  of  this  sort  and  is  glad  to  see 
that  many  architects  are  now  insisting  upon  having  competitive 
designs  thus  checked.  When  they  all  do  it  there  will  undoubtedly 
be  a  radical  change  in  "  common  practice." 

When  an  engineer  is  independently  employed  to  furnish  engi- 
neering service  he  should  not  accept  the  work  if  it  stops  with  the 
furnishing  of  the  plans  and  details.  He  should  insist  upon  being 
retained  as  an  adviser  during  the  progress  of  the  work.  The 
work  of  the  best  designers  may  often  be  discredited  when  the  low- 
est bidder  is  not  possessed  of  enough  experience,  or  honesty,  to 
put  into  the  fabrication  of  the  structure  the  quality  of  work  which 
the  designer  put  into  the  design.  That  many  poorly  designed 
buildings  stand  to-day  is  due  to  the  fact  that  the  work  was  per- 
formed by  honest,  experienced  contractors. 


CHAPTER  II 
Internal  Forces 

THE  term  "  bending  moment "  is  a  description  of  the 
breaking  effect  of  external  forces  on  a  beam.  The  term 
"  resisting  moment  "  is  a  description  of  the  effect  of  internal 
forces  in  a  beam  set  up  to  resist  breaking.  Beams  not  stayed 
laterally  will  bend  to  one  side,  in  which  case  the  full  value  of  the 
resisting  moment  is  not  obtained. 

The  action  of  the  resisting  forces  may  be  illustrated  by  simple 
framework,  for  a  frame  is  merely  a  light  beam  containing  little 
or  no  superfluous  material.  The  ideal  frame  contains  no  superflu- 
ous material,  but  if  this  is  obtained  by  an  increase  in  cost  of  fabri- 
cation the  frame  is  not  ideal  from  the  standpoint  of  the  user, 
regardless  of  the  mathematically  ideal  condition. 

The  designer  soon  finds  that  mathematical  analysis  of  stresses 
treats  frames  as  lines  through  the  center  of  gravity  of  pieces.  It 
is  with  pieces  that  the  designer  has  to  deal.  In  a  mathematical 
design  of  a  frame  all  forces  act  at  points,  and  when  a  line  of  infini- 
tesimal thickness,  dealt  with  by  the  mathematician,  is  replaced  by 
a  piece  of  wood  or  steel  or  concrete,  certain  stresses  are  set  up 
around  the  points.  Then  joints  are  constructed  and  an  Analysis 
must  be  made  of  the  forces  around  the  joints,  this  involving  the 
design  of  rivets  and  fastenings  to  keep  the  joints  from  moving. 

A  simple  beam  contains  some  superfluous  material,  but  much 
of  this  material  acts  to  transmit  stresses  in  various  directions  and 
thus  takes  care  of  internal  stresses  which  otherwise  would  inter- 
fere with  the  direct  action  assumed  to  take  place  along  the  lines 
connecting  the  points  around  which  the  forces  act. 

The  strongest  frame  is  a  triangle,  for  a  triangle  with  sides  of 
fixed  length  cannot  distort  under  direct  stresses  acting  along  the 
center  lines  of  the  pieces  of  which  it  is  composed. 

The  capital  letters  A,  B,  and  C  are  used  to  designate  the  three 
angles  of  a  triangle,  and  the  corresponding  small  letters  are  used 

56 


INTERNAL  FORCES 


57 


to  designate  the  sides  opposite  the  angles.  By  placing  the  capital 
letters  so  small  b  will  represent  the  base  and  small  a  will  represent 
the  altitude  (height)  of  a  triangle  the  student  has  a  mnemonic 
idea  of  the  relations  of  the  sides  and  angles.  In  the  following 

figures 

a  =  BC  (line  from  B  to  C) 
6  =  AC  (line  from  A  to  C) 
c  =  AB  (line  from  A  to  B) 


Fig.  48  —  Frame  with  Inclined  Strut 


Then  in  Fig.  48  b  =  Va2  +  c2  =  8.246, 

u^^-.... 


* ,: zrrr^^ 


Fig.  49  —  Frame  with  Inclined  Strut  and  In- 
clined Tie 

in  Fig.  49       6  =  c  =  V  (a/2)2  +  (line  A-a)2  =  8.062, 


Fig.  50  —  Frame  with  Inclined  Tie 

in  Fig.  50  c  =  Va2  +  b2  =  8.246. 

The  rule  is  known  as  the  "  Rule  of  Pythagoras  "  and  is  given 
in  every  school  arithmetic  as  follows: 

In  every  right-angled  triangle  the  square  on  the  hypothenuse  is 
equal  to  the  sum  of  the  squares  on  the  other  two  sides. 

In  Fig.  48 

cP  =  8  X  3000  =  24,000  ft.  Ibs.  in  member  C. 


58  PRACTICAL  STRUCTURAL  DESIGN 

The  stress  in  c  =  —  =  —7. — •  =  12,000  Ibs.  tension,  for  the  stress 
a  z 

acts  away  from  the  point  B. 

bP  =  8.246  x  3000  =  24,738  ft.  Ibs. 
for  member  6. 

r  r>        cy  A  *7OQ 

The  stress  in  b  =  —  =  — '— —  =  12,369  Ibs.  compression,  for  the 

stress  acts  toward  the  point  C. 

In  Fig.  49  cP  =  bP  =  8.062  x  3000  =  24,186  ft.  Ibs. 

24  186 
Stress  in  c  =  stress  in  6  =  — '- —  =  12,093  Ibs.,  the  character  of 

the  stress  in  each  member  being  determined  by  whether  the  mem- 
ber pulls  from  the  point  of  fastening  or  pushes  toward  the  support. 
In    Fig.   50    cP  =  8.246  x  3000  =  24,738    ft.    Ibs.      Stress    in 

=  12,369  Ibs.  tension. 
bP  =  8  x  3000  =  24,000  ft.  Ibs. 

bP       24  000 
Stress  in  b  =  -^  =      ^      =  12,000  Ibs.  compression. 

The  examples  show  that  the  stress  at  any  point  in  a  horizontal 
member  of  a  frame  is  equal  to  the  bending  moment  at  the  point  divided 
by  the  depth  of  the  frame  at  that  point.  The  stress  in  an  inclined 
member  is  equal  to  the  stress  in  the  corresponding  horizontal  member 
times  the  ratio  of  their  respective  lengths. 

A  frame  is  so  made  that  certain  members  are  hi  tension  and 
other  members  are  in  compression,  the  shear  being  carried  by  the 
vertical  and  inclined  members.  The  lines  of  travel  of  the  stresses 
are  plainly  exhibited.  The  same  lines  exist  in  a  solid  beam,  so  in 
a  beam  it  is  also  true  that  the  horizontal  stress  is  equal  to  the 
bending  moment  divided  by  the  depth  of  the  beam.  On  one  side 
of  the  neutral  axis  the  stress  is  tension  and  on  the  other  side  the 
stress  is  compression.  This  will  be  fully  explained  later. 

Assume  that  the  frame  is  made  of  some  material,  'wood  for 
.example,  in  which  a  fiber  stress  of  500  Ibs.  per  square  inch  can  be 
used  in  either  tension  or  compression.  Referring  to  Fig.  49,  where 
the  stress  in  each  member  is  equal,  each  member  will  require  an 

area  =     cc'nn     =  24.186  sq.  ins.     Extracting  the  square  root  gives 

OUU 

the  dimensions  of  each  piece  as  5.85  ins.  x  5.85  ins.,  which,  of  course, 


INTERNAL  FORCES  59 

will  be  commercial  6  ins.  x  6  ins.  wood,  after  surfacing.  The  pieces 
therefore  can  be  6  ins.  square.  The  lines  of  stress  pass  through 
the  centers  of  the  pieces,  otherwise  some  twisting  and  bending 
strains  will  be  set  up.  Twisting  is  called  torsion  in  such  cases. 

The  actual  construction  of  such  a  bracket  frame  is  shown  in 
Fig.  51,  the  stresses  for  which  are  given  in  Fig.  50.    The  lower 
piece  rests  on  an  angle  at  the  wall  and  a  plate  may  be  placed 
between  the  end   and 
the  wall  if  the  pressure 
exerted  is  greater  than 
the  wall  can  stand. 
The  area  of  the  plate 

will  be  such  that  the        </*-  /A  , 

•„  ,      v  .  •          i*  v,    i^oW**- 

pressure  will  be  distri- 
buted to  an  extent  cal-    Fig.  51  —  Design  for  Sidewalk  Canopy     AN 
culated    to    keep    the 

allowable  compressive  load  on  the  wall  within  proper  limits.  The 
load  P  represents  the  reaction  at  the  outer  end  of  the  frame, 
the  reaction  at  the  wall  end  being  carried  on  the  angle  support. 
The  diagonal  is  a  rod  which  will  be  about  1  in.  in  diameter  if  of 
steel,  for  steel  can  be  stressed  to  16,000  Ibs.  per  square  inch  in 
tension.  To  anchor  the  rod  in  the  wall  a  bolt  1  in.  in  diameter 
extends  into  the  far  side  and  there  a  plate  is  fixed  or  it  may  be 
anchored  in  a  concrete  block  in  the  wall. 

The  circumference  of  a  circle  is  3.1416  x  the  diameter.  The 
circumference  of  a  1-in.  rod  is  3.1416  ins.,  so  for  each  inch  in  length 
there  is  an  area  of  3.1416  sq.  ins.  An  adhesion  of  concrete  to 
steel  of  75  Ib.  per  square  inch  is  customary,  so  the  adhesion  per 
inch  of  length  of  the  anchor  rod  in  the  concrete  =  3.14  x  75  =  236 

12  QCO 
Ibs.    The  total  length  of  the  rod  =    ^    =  52.41  ins.     Instead  of 

using  one  straight  rod  with  a  ring  in  the  end  it  can  be  made 
U-shaped,  each  leg  embedded  in  the  concrete  27  ins.  A  much 
lower  stress  may  be  obtained  by  running  the  tie  rod  higher,  an 
angle  of  about  45  degrees  being  good. 

This  example  was  worked  to  illustrate  the  simplicity  of  such 
computations  and  to  show  that  all  lines  of  stress  must  pass  through 
points.  In  Fig.  51  the  tie  rod  is  shown  to  run  quite  a  distance 
into  the  bottom  member  in  order  to  have  all  the  forces  acting 
property.  In  practical  work  the  vertical  bolt  suspending  the 


60  PRACTICAL  STRUCTURAL  DESIGN 

load  will  have  an  eye  ring  at  the  top  to  receive  the  tie.  The 
eccentricity  will  not  be  large  enough  to  make  a  great  difference, 
but  even  this  small  amount  can  be  greatly  reduced  by  running  the 
tie  at  a  steeper  angle. 

Assume  that  the  lines  of  stress  are  contained  within  a  beam, 
Fig.  52,  anchored  in  the  wall  in  the  usual  way.  The  size  of  the 
beam  is  proportioned  to  take  care  of  the  tensile,  compressive,  and 
shearing  stresses  and  the  anchorage  is  the  only  item  to  be  now  con- 

sidered.   The  wall  reaction  is 
equa]  t0  the  sum  Of  aii  tne 

loads   on    the   beam,    so    the 
weight  of  the  wall  resting  on 
the  beam  must  be  equal  to,  or 
Fig.  52—  Action  of  Moments  in  Anchor-    exceed,  the  reaction.     Weight 
age  of  Cantilever  Beam 


bottom  surface,  and  as  the  wall  must  be  of  a  definite  width  to  rest 
on  the  beam  a  moment  is  created. 

In  Fig.  52  the  weight  of  the  wall  multiplied  by  the  distance 
from  the  face  to  the  center  of  gravity  must  be  equal  to  the  moment 
obtained  by  multiplying  the  loads  on  the  beam  by  the  respective 
distances  from  their  centers  of  gravity  to  the  face  of  the  wall. 

The  illustration  of  how  the  stresses  are  obtained  is  true  for 
beams  or  frames  resting  on  two  or  more  supports,  the  stress  being 
equal  to  the  moment  at  a  point  divided  by  the  depth  at  that 
point.  When  the  lower  member  is  a  square  piece  of  timber  the 
distance  is  measured  from  its  center  and  when  the  upper  member 
is  a  rod  the  distance  is  measured  to  the  center  of  the  rod.  That  is, 
all  distances  are  measured  between  centers  of  gravity  of  the  parts, 
or  members,  of  a  frame,  the  total  over-all  height  from  the  top  to 
the  bottom  being  equal  to  the  distance  center  to  center  plus  half 
the  thickness  of  each  piece. 

When  a  frame  or  truss  is  composed  of  angles  or  other  rolled 
shapes  the  distance  is  always  measured  between  centers  of  gravity 
of  the  top  and  bottom  chords.  Thus  when  depth  is  mentioned 
it  is  not  the  over-all  depth.  The  stress  obtained  is  the  stress  on 
the  center  line  passing  through  the  center  of  gravity,  the  stress 
being  slightly  larger  at  the  outer  edge  of  the  rolled  section  and 
slightly  smaller  at  the  inner  edge,  when  there  is  bending.  In 
pieces  acting  as  plain  ties  or  plain  struts,  so  .the  stress  is  pure 
tension  or  pure  compression  without  bending  stress,  the  stress  is 


INTERNAL  FORCES  61 

equal  over  the  entire  area.  In  a  frame  pieces  are  generally  so 
placed  that  the  stresses  are  purely  tension  or  compression,  but  a 
beam  is  a  solidly  filled  frame  and  the  stress  is  greatest  at  top  and 
bottom,  reducing  to  zero  on  the  line  where  the  compressive  force 
changes  to  tension.  It  is  necessary  then  to  find  the  position  of 
the  center  of  gravity  of  the  beam  on  each  side  of  the  neutral 
plane. 

The  question  may  be  asked,  "  Why  is  the  depth  used  as  a 
divisor?  "  Referring  to  Fig.  48  and  Fig.  49  a  dotted  line  is  shown 
from  B  to  D.  The  moment  is  obtained  by  multiplying  the  load 
by  the  horizontal  distance  from  the  wall.  To  resist  this  moment, 
which  means  to  prop  up  the  member  AB,  there  must  be  some 
force  exerted  at  a  distance  BD  from  the  point  B.  That  is,  the 
bending  moment  and  the  moment  to  resist  it  are  taken  about  the 
point  B.  The  bending  moment  at  B  =  8  x  3000  =  24,000  ft.  Ibs. 
This  is  resisted  by  some  force  acting  about  the  point  B  with  an 
arm  =  BD.  Thus  the  upward  pushing  force  in  the  member 
AC  must  equal  the  downward  moment  divided  by  the  length  BD. 
This  upward  force  is  a  reaction  and  is  compressive. 

To  obtain  a  reaction  multiply  the  loads  by  the  distance  through 
which  they  act  and  divide  by  the  span  length  between  supports. 
The  obtaining  of  tensile  and  compressive  stresses  is  the  same 
thing.  First  a 'downward  bending  moment  is  obtained  and  then 
a  reaction  is  found  by  dividing,  not  by  the  span  of  the  beam,  but 
by  the  span  between  supports.  There  is  an  upper  support  to 
which  the  tension  member  is  fastened  and  a  lower  support  against 
which  the  compression  member  abuts.  The  distance  between  them 
is  the  span  between  supports.  This  span  is  the  distance  measured 
on  the  shortest  line  between  the  lines  representing  the  direction 
of  the  forces,  so  it  is  perpendicular  to  the  direction  of  the  inclined 
member.  For  all  practical  purposes  the  lower  support  is  at  D 
and  not  at  C.  The  member  is  merely  carried  on  to  the  point  C. 

It  is  correct  to  multiply  the  load  by  the  arm  BA  and  divide 
by  the  arm  BD  in  all  cases,  but  considerable  work  must  be  done 
to  obtain  the  length  of  the  arm  BD.  This  requires  a  knowledge 
of  geometry  and  trigonometry  and  the  use  of  tables  of  functions 
of  angles.  To  obtain  the  length  of  the  inclined  member  and  use 
this  as  a  moment  arm  and  then  divide  by  the  vertical,  distance 
between  the  centers  of  gravity  of  the  top  and  bottom  members, 
is  the  shortest  method  and  commonly  used,  for  the  result  is  correct. 


62 


PRACTICAL  STRUCTURAL  DESIGN 


The  experienced  structural  designer  uses  tables  of  squares  to 
lighten  the  labor  of  making  the  computations  involved  in  obtaining 
the  hypothenuse  of  a  triangle.  Barlow's  Tables  contain  the 
squares,  cubes,  square  roots  and  cube  roots  of  all  numbers  from 
1  to  10,000.  Smoley's  Tables  for  the  Use  of  Structural  Designers 
contain  the  squares  of  all  lengths  up  to  100  ft.,  varying  by  six- 
teenths of  an  inch. 

Moment  of  Resistance 

In  Fig.  53  is  illustrated  a  beam  loaded  so  there  is  a  tendency 
to  bend,  which  tendency  is  resisted  by  compression  in  the  upper 

fibers  and  tension  in 
the  lower  fibers. 
The  beam  rests  on 
-»j         two  supports.    If  it 
were  a  cantilever 
Fig.  53  —  Compressive  and  Tensile  Force  Triangles    keam   the   compres- 
in  a  Beam 

sion    would    be    in 

the  lower  fibers  and  the  tension  in  the  upper  fibers.  The  two 
triangles  show  graphically  the  forces,  the  horizontal  shading  repre- 
senting the  compressive  force  and  the  vertical  shading  represent- 
ing the  tensile  force.  The  overlapping  portions  of  the  triangles 
neutralize  each  other,  and  the  force  triangles  in  Fig.  54  are  the 
result.  In  the  case  considered  the  beam  is  uniform  hi  cross  sec- 
tion and  the  stresses  in  tension  and  compression  are  equal ;  there- 
fore the  triangles  are  equal  in  size  and  the  neutral  plane  where 
the  triangles  meet  is  in  the  center  of  area  of  the  cross  section. 

The  neutral  plane 
is    a    plane    in   the    t*~ 
center  of  gravity  of 
the  section,  parallel    | 
with  the  upper  and 
lower     surfaces, 
where    there    is    no    Fig.  54 -The  Usual Method  of Representing  Action 

of  Resisting  1  orces  in  a  Beam 
tension  or  compres- 
sion.    The  stresses  above  and   below   are   opposite  in  nature; 
therefore  the  only  stress  along  the  neutral   plane  is  horizontal 
shear.     The  word  "  neutral"  implies  that  in  this  plane  opposite 
forces  are  neutralized. 

This  definition  of  the  neutral  plane  is  important.  In  all  text 
books  the  illustrations  refer  to  beam  sections  of  homogeneous 


i 


INTERNAL  FORCES  63 

material,  with  equal  tensile  and  compressive  unit  stresses.  There- 
fore the  center  of  gravity  of  the  beam  section  is  the  center  of 
gravity  of  the  area  of  the  section.  For  this  reason  a  great  many 
students  claim  that  the  neutral  plane  is  always  in  the  center  of 
gravity,  which  is  correct,  but  they  insist  that  this  section  of 
gravity  is  at  mid-depth,  which  is  not  always  correct. 

The  total  tension  must  equal  the  total  compression.  When  a 
beam  is  strong  in  compression  and  weak  in  tension,  or  vice  versa, 
the  neutral  plane  will  not  be  in  the  center  of  area  of  the  cross 
section  if  the  beam  is  not  proportioned  for  this  condition,  but  it 
will  be  in  the  center  of  gravity.  We  will  now  explain  this  state- 
ment, which  appears  to  be  contradictory.  Assume  the  stress 
triangles  in  Fig.  54  to  be  equal,  although  the  tensile  fiber  stress 
may  be  lower  than  the  compressive  fiber  stress.  The  fact  that 
the  two  forces  must  balance  makes  the  height  of  one  triangle 
greater  than  the  other,  for  "  fiber  stresses  are  directly  propor- 
tional to  their  distance  from  the  neutral  axis."  We  therefore  have 
two  triangles  of  stress,  one  with  a  wide  base  (high  fiber  stress) 
and  the  other  with  a  narrow  base  (low  fiber  stress).  The  heights 
obviously  must  vary  in  proportion  in  order  that  these  "  stress 
areas  "  will  be  equal. 

The  illustrations  are  of  an  imaginary  beam  section,  a  thin  slice, 
so  the  position  of  the  neutral  plane  is  represented  by  a  line  called 
the  "  neutral  axis  "  and  the  triangles  represent  the  sides  of  wedges 
across  the  beam.  Much  of  the  difficulty  met  with  by  students  in 
regard  to  the  neutral  axis  arises  from  the  fact  that  such  illustra- 
tions are  used.  The  material  is  not  actually  stretched  and  short- 
ened as  indicated,  the  triangles  being  imaginary,  each  with  a 
base  representing  a  force  and  not  a  length. 

The  neutral  axis  is  in  the  center  of  gravity  of  a  section,  sym- 
metrical or  unsymmetrical,  provided  such  position  is  the  center 
of  gravity  of  a  couple  in  the  section.  If  the  stresses  are  different 
but  the  section  is  symmetrical  the  neutral  axis  will  be  nearer  the 
side  having  the  higher  stress.  If  the  section  is  unsymmetrical, 
to  balance  the  difference  in  the  fiber  stresses,  the  neutral  axis  will 
be  in  the  center  of  gravity  of  the  section  and  also  in  the  center 
of  area.  An  example  of  this  is  the  cast  iron  beam.  Before  taking 
it  up  a  definition  will  be  given  of  a  couple.  A  couple  consists  of 
two  equal  opposite  forces  acting  in  parallel  lines  about  a  point. 
One  illustration  is  a  load  on  a  beam  and  the  reaction  at  the  end 


64  PRACTICAL  STRUCTURAL  DESIGN 

of  the  beam.  Another  is  the  tensile  force  in  a  beam  balanced  by 
the  compressive  force.  They  are  equal  and  opposite  in  amount, 
acting  at  the  end  of  an  arm  measuring  the  distance  between  the 
centers  of  gravity  of  the  two  forces. 

It  is  customary  to  use  in  cast  iron  a  tensile  fiber  stress  of  3000 
Ibs.  per  square  inch  and  a  compressive  fiber  stress  of  10,000  Ibs. 
per  square  inch. 

A  cast  iron  beam,  therefore,  is  made  in  the  form  of  an  inverted 
T,  the  well-known  cast  iron  window  lintel  being  an  example  in 
point.  First  the  beam  was  proportioned  and  the  center  of  gravity 
of  the  cross  section  found.  The  tensile  stress  being  the  maximum 
at  the  lower  surface,  the  average  stress  multiplied  by  the  area  of 
the  section  below  the  center  of  gravity  gave  the  tensile  force 
(total  tensile  stress).  The  area  above  the  center  of  gravity  tunes 
one-half  the  maximum  compressive  fiber  stress  gave  the  total 
compressive  force.  If  they  were  not  equal  a  new  section  would 
be  chosen  and  after  a  few  trials  a  section  would  be  obtained  in 
which  the  area  below  the  neutral  axis  tunes  the  average  tensile 
fiber  stress  equaled  the  area  above  the  neutral  axis  times  the 
average  compressive  fiber  stress. 

This  particular  example  is  interesting  because  it  disproves  a 
statement  frequently  met  with  in  books  of  a  certain  class,  namely 
that  "  the  moments  of  the  horizontal  forces  on  the  two  sides  of 
the  neutral  axis  must  be  equal."  Assuming  this  statement  to  be 
true,  the  distance  from  the  center  of  gravity  of  the  tensile  area  to 
the  neutral  axis  must  equal  the  distance  from  the  center  of  gravity 
of  the  compressive  area  to  the  neutral  axis.  Having  located  the 
position  of  the  neutral  axis  as  above  described,  take  a  moment 
arm  from  the  neutral  axis  to  the  center  of  area  of  each  section  and 
multiply  the  area  by  the  average  stress  times  the  moment  arm. 
One  trial  will  show  the  falsity  of  the  statement.  The  force  (stress) 
areas  on  each  side  must  be  equal,  and  the  moments  do  not  balance 
about  the  neutral  axis,  except  when  the  cross  section  is  sym- 
metrical and  the  tensile  fiber  stress  is  equal  to  the  compressive 
fiber  stress.  The  moment  arm  is  measured  from  the  center  of 
gravity  of  the  stress  triangle  on  one  side,  not  the  center  of  area, 
to  the  center  of  gravity  of  the  stress  triangle  on  the  other  side. 
The  moments  of  resistance  will  be  equal,  which  is  quite  a  different 
statement  from  that  which  makes  the  moments  equal  on  the 
two  sides  of  the  neutral  axis. 


INTERNAL  FORCES  65 

In  Fig.  54  let  the  small  /  stand  for  the  maximum  fiber  stress,  that 
is  the  unit  stress,  which  is  generally  expressed  in  pounds  per  square 
inch.  Then/c  =  unit  compressive  stress  and  ft  =  unit  tensile  stress. 
When  the  stresses  are  equal  the  letter  /  is  used  without  a  subscript. 

All  forces  must  act  through  the  center  of  gravity  of  bodies  or 
areas  in  order  to  effect  a  movement  of  the  whole  without  turning 
it,  as  about  an  axis.  The  center  of  gravity  of  a  triangle  is  one- 
third  the  distance  from  the  base.  In  Fig.  54  the  triangle  on  one 
side  of  the  neutral  axis  has  a  height  equal  to  one-half  the  total 

height  of  the  beam,  therefore  is  represented  by  -•    The  width  of 

z 

the  triangle  at  the  base  is  represented  by  the  unit  stress  at  that 
place,  /.  The  triangle  is  a  force  triangle  and  the  area  is  equal  to 
the  total  force  exerted  on  it.  This  is  found  by  the  ordinary  rule 
for  areas  of  triangles  —  half  the  base  multiplied  by  the  height  —  or 

A-f*h-fh 
~  2  X  2  -  T 

The  length  j  is  known  as  the  moment  arm.  The  moment  arm 
is  the  distance  between  the  centers  of  gravity  of  the  tension  and 
compression  members.  The  lower  triangle  is  the  tension  member, 
if  we  assume  the  beam  to  be  a  frame,  and  the  upper  triangle  is 
the  compression  member,  the  neutral  plane  being  the  dividing 
line.  Since  the  forces  act  at  a  point  and  this  point  is  the  center 
of  gravity  of  the  member,  then  the  total  force  in  compression, 

—,  acts  to  balance  the  total  force  in  tension,  ~,  with  a  moment 

2      h      4/i      2/i 
arm  =  2  ><  3  X  2  =  T  =  T 

The  moment  arm  times  the  compressive  (or  tensile)  force  gives 
the  moment  of  resistance  per  unit  of  breadth, 

2h       h2VV 


This  reasoning  has  been  based  on  a  breadth  equal  to  1,  or  unity. 
To  make  practical  use  of  the  expression  the  breadth,  designated 
by  6,  must  be  introduced.  This  gives  the  expression  for  the 
moment  of  resistance  of  a  beam  of  homogeneous  material  and 
rectangular  cross  section  commonly  seen  in  text  books, 

M  -fbh*- 

MT---' 


66  PRACTICAL  STRUCTURAL  DESIGN 

Some  writers  use  d  (depth)  instead  of  h  (height),  but  as  d  is 
used  these  days  to  indicate  the  depth  from  the  top  of  a  reinf  orced- 
concrete  beam  to  the  center  of  gravity  of  'the  steel  reinforcement 
(the  real  depth  of  a  reinforced-concrete  beam)  the  letter  h  is 
preferred  when  the  total  over-all  height  or  depth  of  a  beam  is 
meant.  In  a  reinforced-concrete  beam  the  concrete  below  the 
center  of  the  steel  is  used  solely  for  bond  and  protection  and  is 
not  considered  in  computations  to  ascertain  the  strength  of  the 
beam. 

The  moment  of  resistance  of  a  rectangular  beam  of  homogeneous 
material  is  said  to  be  "  the  section  modulus  times  the  fiber  stress." 
This  means  that 


is  an  expression  denoting  the  effect  of  the  shape  of  the  beam  or  the 
resistance  it  offers  to  destruction  by  loading.  No  matter  what 
the  material  or  what  the  stress  used,  the  effect  of  the  shape  is  the 
same  and  this  is  called  the  "  section  modulus,"  the  word  "  modulus" 
meaning  "  measure."  It  is,  therefore,  the  measure  of  the  resistance 
of  the  shape. 

Every  shape  has  a  section  modulus  designated  by  S  in  the 
steel  manufacturers'  handbooks.  The  calculation  of  the  section 
modulus  for  a  beam  with  rectangular  cross  section  has  been 
given,  as  it  is  the  most  simple  section  to  handle  without  confusing 
the  reader  with  a  mass  of  figures. 

To  compute  the  section  modulus  for  any  shape  first  assume 
some  axis  passing  through  the  center  of  gravity.  Then  divide 
the  area  into  any  number  of  layers  desired  by  lines  parallel  to  the" 
axis.  Find  the  area  of  each  layer  and  the  distance  from  the  axis 
to  the  center  of  gravity  of  each  layer.  Multiply  each  layer  by 
the  square  of  the  distance  from  the  axis  to  the  center  of  gravity 
of  the  layer  and  add  the  products.  Thus  is  obtained  the  Moment 
of  Inertia,  an  expression  denoting  the  disinclination  of  the  body 
to  move  as  a  whole.  The  Moment  of  Inertia  divided  by  the  distance 
from  the  axis  to  the  highest  stressed  fiber  gives  the  Section 
Modulus. 

For  a  rectangular  beam  the  moment  of  inertia  is 


12' 


INTERNAL  FORCES  67 

and  this  divided  by  the  distance  from  the  neutral  axis  to  the 
most  distant  fiber  gives  the  section  modulus,  as  follows: 
W_h  =  bW      2  =  fe^ 
12   :  2       12  X  n       6 
the  distance  from  the  neutral  axis  to  the  skin  being  h  +  2. 

The  section  modulus  is  dependent  entirely  upon  the  shape  and 
is  independent  of  the  weight  of  a  beam  and  of  the  strength  of  the 
material  in  the  beam.  Tables  giving  the  section  modulus  when 
once  computed  are  good  for  all  time.  In  steel  manufacturers' 
handbooks  tables  are  given  of  the  section  moduli  for  every  shape 
rolled,  so  the  proper  beam  may  be  selected  when  the  bending 
moment  is  known  and  the  fiber  stress  is  known. 
Let  M  =  moment  (bending  moment  =  resisting  moment)  in  inch 

pounds. 

S  =  section  modulus  in  inches. 

/  =  allowable  maximum  fiber  stress  in  pounds  per  square 
inch. 

then    M  =  SfandS  =  j- 

Me 

In  many  books  the  expression  /  =  ^y-  is  encountered.    The 

moment  divided  by  the  section  modulus  gives  the  fiber  stress.    The 

section  modulus  =  -.  in  which 
c 

I  =  moment  of  inertia. 

c  =  distance  from  the  neutral  axis  to  the  most  stressed  fiber. 
Sometimes  y  is  used  instead  of  c.  -  _, 

,      M       Me 
Therefore  /  =  — =  _ 

One  method  for  finding  the  Moment  of  Inertia  and  the  Section 
Modulus  for  T-sections,  L's,  etc.,  is  to  first  assume  a  rectangular 
section  having  dimensions  equal  to  the  extreme  outside  dimen- 
sions of  the  shape.  Find  the  properties  (i.e.,  I  and  S)  for  this 
rectangular  section.  Next  take  each  hollow  portion  considered 
as  a  smaller  rectangular  section  and  find  the  properties.  Adding 
the  results  for  each  of  the  pieces  cut  away  and  subtracting  the 
sum  from  the  properties  for  the  entire  section,  the  properties  are 
found  for  the  remainder. 

Example.  —  What  is  the  section  modulus  for  a  hollow  rec- 
tangular section  having  an  outside  width  of  8  ins.  and  an  outside 


68  PRACTICAL  STRUCTURAL  DESIGN 

depth  of  12  ins.,  the  thickness  of  the  shell  being  1  in.?    Axis 
horizontal. 

S  (for  entire  section)    =  f  =  8x12x12  -  192 

8  (for  interior  section)  ,  6|  =  «  x  10  x  10  .  100 

S  (for  metal  section  of  hollow  shape)  =    92  his. 

Example.  —  What  is  the  section  modulus  for  a  T-section  12  ins. 
deep  over-all  with  an  extreme  width  of  8  ins.  and  with  stem  and 
flanges  each  \  in.  thick?  Axis  horizontal. 

,      6/i2      8  x  12  x  12 
S  (for  entire  section)  =  —  =  ^ =  192  ins. 

S  (for  section  on  one  side)  =  3'75  ***     —  =  75.625  ins.,   and 
2  X  75.625  =  151.25  ins. 

The  S  for  the  section  =  192  -  151.25  =  40.75  ins.  The  fiber 
stress  is  called  by  some  writers  the  "  skin  stress,"  a  very  good 
term,  for  it  is  actually  the  stress  in  the  outer  skin,  which  is  assumed 
to  have  no  thickness,  or  has  an  infinitesimal  thickness.  The 
stress,  within  the  elastic  limit,  varies  uniformly  as  a  straight  line 
to  zero  at  the  neutral  axis.  Therefore  on  each  layer  between  the 
skin  and  the  neutral  axis  the  stress  is  less  than  that  assumed 
in  the  computations.  The  total  stress  is  equal  to  the  average 
stress  multiplied  by  the  distance  from  the  neutral  axis  to  the 
skin. 

In  Fig.  55  two  beam  sections  are  shown  with  the  axes  at  right 
angles  and  the  respective  moments  of  inertia  and  section  moduli 
are  also  given.  The  moment  of  resistance  depends  upon  the  square 
of  the  depth,  so  that  for  two  rectangular  beams  of  homogeneous 
material,  having  the  same  breadth,  the  beam  having  a  depth 
twice  as  great  as  that  of  the  other  beam  has  a  resisting  moment 
four  times  as  great.  It  will  also  be  much  stiffer,  so  there  will 
be  less  deflection  with  a  deep  beam.  The  most  economical 
beam,  considering  stiffness  and  strength,  with  a  rectangular 
cross  section,  has  a  breadth  between  two-thirds  and  three-fourths 
the  depth. 

A  beam  of  /-section  is  possible  in  steel  and  iron  because  of  the 


INTERNAL  FORCES 


69 


L 


Ax/sA-A    • 

1-720  1-2/5.8 

S-120  S-  3&0 

AxisB-B 
1-125  1-95 


strength  of  these  materials.  The  material  is  so  disposed  that 
practically  all  the  metal  highly  stressed  is  concentrated  in  the 
flanges,  the  web  transmitting  the  stresses  and  taking  care  of 
shear.  When  a  beam  of  /-section  is  required  having  a  depth  greater 
than  can  be  properly  rolled,  one  is  made  of  a  plate  having  angles 
riveted  along  the  edges,  this  being  known  as  a  plate  girder.  When 
a  still  deeper  girder  is  required  a  lat- 
ticed girder  is  used,  this  being  known 
as  a  truss. 

Wooden  beams  are  made  only  in 
solid  form,  rectangular  or  round,  for 
wood  is  composed  of  distinct  fibers, 
many  of  which  would  be  completely 
detached  from  the  main  fibers  in 
shaping  the  section  to  provide  broad 
flanges.  Steel  and  iron  will  transmit 
stresses  equally  well  in  all  directions, 
so  while  in  the  filleted  section  con- 
necting the  flange  to  the  web  there  is  I  ig.  55  —  Comparison  of  Mo- 
some  concentration  of  stresses,  this 
has  been  taken  care  of  in  designing 
the  beam.  In  all  solid  and  rolled 
shapes  the  maximum  fiber  stress  is  the  skin  stress.  In  built-up 
sections,  such  as  plate  or  latticed  girders,  the  maximum  fiber 
stress  is  assumed  to  cover  the  flange  member  and  the  whole 
action  is  on  the  line  passing  through  the  center  of  gravity  of  the 
flange.  The  stress  is  transmitted  from  the  web,  or  the  web 
members,  to  the  flange  through  rivets,  which  must  be  properly 
proportioned  in  size  and  properly  spaced  to  take  care  of  the  shear. 

Elastic  Limit 

Within  the  limit  of  strength  known  as  the  "  elastic  limit,"  all 
materials  may  be  stressed  a  number  of  times  and  recover  their 
original  dimensions.  The  elastic  limit  is  a  stress  where  the 
material  is  permanently  deformed  and  stress  in  excess  of  the 
elastic  limit  causes  rapid  deformation.  Up  to  the  elastic  limit 
the  stress-strain  curve  is  straight,  but  it  curves  after  the  elastic 
limit  is  passed.  Fig.  56  is  a  stress-strain  diagram  of  steel,  iron, 
and  wrood  and  gives  a  good  idea  of  the  relative  strengths  and 
deformations.  The  illustration  is  copied  from  "Materials  of 


ments  of  Inertia  and  Section 

Moduli  in  a  Rectangular 

Beam  and  in  an  I-beam 


70 


PRACTICAL  STRUCTURAL  DESIGN 


Construction,"  by  the  late  Professor  J.  B.  Johnson.  His  descrip- 
tion of  the  apparent  elastic  limit  is  that  when  it  is  reached  the 
deformation  for  each  increment  of  stress  is  about  double  the 
deformation  for  the  increment  immediately  preceding.  The  curve 
becomes  a  parabola,  or  is  very  nearly  parabolic. 

All  the  statements  made  about  the  moment  of  resistance  of 
beams  are  true  only  within  the  elastic  limit  of  the  material,  for 

they  are  based  on 
"Hooke's  Law"  that 
"  Stress  is  proportional 
to  strain."  When  the 
elastic  limit  is  reached 
the  law  is  no  longer 
true.  A  generation  ago 
when  steel  was  stressed 
to  16,000  Ibs.  per  square 
inch  it  was  said  to  have 
a  factor  of.  safety  of  4, 
based  on  the  ultimate 
strength  of  the  steel, 
64,000  Ibs.  per  square 
inch.  To-day  the  fac- 
tor of  safety  is  based 
on  the  elastic  limit,  and 
as  this  varies  between 


Fig.  56  —  Stress-strain  Diagram  of  Steel,  Iron, 
and  Wood 


29,000  and  36,000  Ibs.  per  square  inch,  averaging  about  32,000  Ibs., 
depending  upon  the  hardness  of  the  steel,  the  factor  of  safety  is 
said  to  be  2,  based  on  the  elastic  limit.  There  is  a  permanent  set 
after  the  elastic  limit  is  passed  and  a  progressive  weakening,  even 
though  the  material  may  not  fail  until  it  is  stressed  up  to  four 
times  the  allowable  working  fiber  stress. 

Modulus  of  Elasticity 

The  modulus  of  elasticity  is  a  number  obtained  by  dividing  the 
stress  by  the  deformation  it  causes.  If  it  were  a  force  it  would 
be  defined  as  a  force  which  will  stretch  a  unit  piece  of  material  to 
twice  its  length  or  compress  it  one-half.  It  is  not  a  force,  nor  is 
it  a  pure  number,  for  it  is  expressed  in  pounds. 

A  certain  steel  bar  having  an  area  of  one  square  inch  was  stressed 
in  tension  16,000  Ibs.  and  the  stretch  carefully  measured  was 


INTERNAL  FORCES  71 

found  to  be  0.000534  the  length.     What  was  the  modulus  of 

elasticity? 

J^ffff,  =  30,000,000  Ibs.  (in  round  numbers). 

U.UUUDO4: 

In  the  example  the  bar  was  one  inch  square.  It  may  make  it 
more  clear  if  the  modulus  of  elasticity  is  defined  as  the  ratio  found 
by  dividing  the  unit  stress  by  the  unit  deformation,  or  unit  strain. 
Stress  is  a  force  and  strain  is  the  deformation  produced  by  a  force. 

The  modulus  of  elasticity  is  used  to  compare  the  relative 
deformation  of  materials  which  must  act  together.  Steel  may  be 
said  to  have  a  modulus  of  elasticity  of  30,000,000.  Concrete  is 
made  of  so  many  different  mixtures  and  the  workmanship  varies 
so  greatly  between  specimens  that  an  average  value  of  the  modulus 
of  elasticity  must  be  taken  for  each  mixture.  The  average  value 
of  1:2:4  concrete  is  2,000,000  and  the  ratio  between  the  moduli 
of  elasticity  of  structural  grade  steel  and  1:2:4  concrete  is  taken 

30  000  000 
to  be    2  OOQ  OOQ  =  15-     The  writer  a  few  years  aS°  caUed  this 

the  "  ratio  of  deformation  "  instead  of  the  "  ratio  between  moduli 
of  elasticity,"  and  his  term  is  very  commonly  used  now.  To  give 
a  clear  explanation  of  the  matter  assume  a  piece  of  concrete  with 
a  unit  cross-sectional  area  and  beside  it  a  piece  of  steel  with  the 
same  area.  An  equal  load  is  placed  on  each  piece  and  the  short- 
ening measured.  It  will  be  discovered  that  the  steel  shortened 
one-fifteenth  as  much  as  the  concrete,  therefore  the  ratio  of  defor- 
mation =  15.  Concrete  can  really  have  no  modulus  of  elasticity,  for 
it  is  a  brittle  material  with  an  elastic  limit  very  difficult  to  measure. 
When,  however,  concrete  is  tested  it  shows  enough  consistency 
in  deformation  to  warrant  the  adoption  of  a  value  for  the  modulus 
of  elasticity  by  means  of  which  a  workable  ratio  of  deformation 
may  be  obtained. 

Reinforced  Concrete  Beams 

Whereas  in  beams  of  a  uniform  material  there  is  a  gradual 
and  uniform  increase  in  stress  from  the  neutral  axis  to  the  top 
and  bottom,  in  beams  of  reinforced  concrete  this  is  true  only  of 
the  upper  portion  of  the  beam.  Roughly,  the  tensile  strength  of 
concrete  is  about  one-tenth  the  compressive  strength.  Since  in  a 
beam  the  tensile  and  compressive  forces  must  be  equal,  the  neutral 
axis  in  a  beam  of  plain  concrete  under  load  will  be  very  high. 


72 


PRACTICAL  STRUCTURAL  DESIGN 


The  tensile  stress  is  low  and  the  eompressive  stress  is  high,  so  the 
height  of  the  tensile  force  triangle  will  be  greater  than  the  height 
of  the  eompressive  force  triangle.  Steel  is  placed  near  the  bot- 
tom of  a  reinforced  concrete  beam  to  give  it  increased  tensile 
strength,  and  this  lowers  the  neutral  axis.  The  concrete  is  not 
relied  upon  to  furnish  any  tensile  strength,  this  being  concen- 
trated in  the  steel.  The  concrete  below  the  neutral  axis  is  there- 
fore used  only  to  furnish  shearing  strength  and  protect  the  steel 
from  corrosion. 

The  total  tensile  stress  is  considered  as  being  carried  by  the 
steel,  and  the  eompressive  stress  is  carried  by  the  concrete  above 
the  neutral  axis,  where  the  variation  in  fiber  stress  follows  the 


.  n      5 


Fig.  57 

straight  line  law  already  considered.  Actually,  the  eompressive 
stress  varies  as  a  parabola  and  not  as  a  triangle,  but  to  use  the 
triangle  is  safe  and  the  "  straight-line  method  "  alone  is  permitted 
in  building  ordinances  and  in  all  regulations  issued  by  responsible 
officials  in  this  and  other  countries. 

To  properly  treat  reinforced  concrete  design  will  require  a  book, 
and  the  author  has  one  in  preparation  which  will  shortly  follow 
the  present  book,  and  replace  his  "  Reinforced  Concrete,  a  Manual 
of  Practice,"  written  in  1907  and  now  out  of  print.  The  method 
he  uses  for  determining  the  position  of  the  neutral  axis  in  a  rein- 
forced concrete  will  be  given  here,  together  with  his  method  for 
determining  the  percentage  of  steel.  The  methods,  and  the 
resulting  formulas,  he  believes  to  be  original,  as  they  have  never 
appeared  in  any  book  to  his  knowledge,  or  to  the  knowledge  of  a 
large  number  of  teachers  and  consulting  engineers  to  whom  he 
wrote  about  the  matter. 


INTERNAL  FORCES  73 

We  have  in  reinforced  concrete  a  material  (steel)  stressed  in 
tension  which  is  from  20  to  30  times  stronger  than  the  compressive 
strength  of  the  concrete.  Furthermore,  the  weaker  material  has 
a  deformation  under  load  15  tunes  greater  than  the  deformation 
of  the  stronger  material. 

Let  E,  -  modulus  of  elasticity  of  the  steel. 
Ec  =  modulus  of  elasticity  of  the  concrete. 

n  =  ration  of  deformation  =  -=?•  (usually  15). 

fs  =  allowable  fiber  stress  (working  stress)  in  the  steel. 
fe  =  allowable  fiber  stress  (working  stress)  in  the  concrete. 

m  =  stress  ratio  =  -r* 

d  =  depth  from  top  of  concrete  beam  to  center  of  gravity 

of  the  steel  reinforcement. 
k  =  depth  from  the  top  of  the  beam  to  the  neutral  axis 

(expressed  as  a  percentage  of  d). 

j  =  moment  arm  expressed  as  a  per  cent  of  d.    When  the 
value  of  d  is  given  in  inches  then  the  moment  arm  is  jd. 

k  dk 

j  =  1  -  -  and  jd  =  d  -  —- 

Referring  to  Fig.  57;  on  a  piece  of  squared  paper  set  off  ten  units 
vertically.  Measuring  to  the  right  along  the  top  lay  off  n.  Measur- 
ing to  the  right  along  the  bottom  lay  off  m.  Connect  the  ends  as 
shown  to  form  two  triangles  which  cross  at  the  neutral  axis.  The 
depth,  k,  to  the  neutral  axis  may  then  be  measured  on  the  paper. 
The  two  triangles  show  an  obvious  geome- 
trical relation  which  enables  us  to  find  k  j- 
by  computation,  thus:  i  ~» 

n 


n  +  m 

Fig.  58  illustrates  the  triangle  of  com- 
pressive force.     The  area  of  the  triangle 

f 
=  J-^xk.    The  tensile  force  is  equal  to  the  Fig  58 

area  of  the  steel  times  the  fiber  stress,  so  the  area  of  the  steel  will 
be  obtained  by  dividing  the  compressive  force  by  the  steel  fiber 
stress.  Units  are  used  throughout,  so  the  steel  area  will  be  the 
ratio  between  the  concrete  and  the  steel.  Then, 


74  PRACTICAL  STRUCTURAL  DESIGN 


/•  fc 

Another  expression  for  p  is  found  as  follows  :  The  unit  moment 
of  resistance  is  expressed  by  a  Resistance  Factor,  R.  For  the  steel 
this  is  Rs  and  for  the  concrete  this  is  Re. 


kj  and  R,  =  pjf.. 


The  moment  of  resistance  of  a  concrete  beam  =  Rbd?,  in  which 
R  is  the  resistance  factor  for  the  "  balanced  "  beam,  that  is,  a 
beam  in  which  the  tensile  force  exactly  equals  the  compressive 
force.  Fig.  59  is  a  chart  for  obtaining  R  and  p  for  different 
stresses  in  steel  and  concrete.  The  stresses  recommended  by 
the  Joint  Committee  on  Concrete  and  Reinforced  Concrete  are 
16,000  Ibs.  per  square  inch  for  the  steel  and  650  Ibs.  per  square 
inch  for  1:2:4  concrete. 

For  rectangular  reinforced-concrete  beams  (and  for  slabs  with 
width  6,  of  12  ins.)  the  following  formulas  are  used,  the  moment 
being  in  inch  pounds  and  all  beam  dimensions  in  inches. 

M  =  Rbd2 
M 

Hi 

Rb 

When  the  moment  is  in  foot  pounds,  b  will  be  in  feet  and  d 
in  inches,  or  6  and  d  will  be  in  inches  and  R  will  be  divided 
by  12. 

The  Portland  Cement  Association,  Chicago,  111.,  is  an  organiza- 
tion supported  by  the  cement  manufacturers  of  the  United  States 
and  Canada  for  the  purpose  of  disseminating  information  about 
cement  and  concrete.  Every  man  in  the  building  business  should 
have  his  name  and  address  on  the  mailing  list,  for  some  of  the 
Association  bulletins  deal  with  questions  of  design,  while  all  the 
bulletins  should  be  on  file  in  the  office  of  every  one  who  has  anything 
to  do  with  construction  work. 


INTERNAL  FORCES 


75 


0.10          0. 1 5>      0.20  0.25  0.30  035  (MO  0.45  0.30  C.60  0.70  0.80  0.90 1.0 

Percentage  of  Reinforce  men  I 
Fig.  59  —  Coefficients  of  Resistance  of  Reinforced  Concrete  Beams 


76  PRACTICAL  STRUCTURAL  DESIGN 

Shearing  Resistance 

A  beam  may  be  strong  enough  to  carry  the  load  without  a 
bending  failure,  which  will  crush  the  fibers  at  the  top  or  pull  them 
apart  at  the  bottom,  yet  it  may  fail  in  shear.  The  direct  shearing 
stress  at  any  section  on  a  beam  is  found  by  dividing  the  shear  at 
the  section  by  the  area  of  the  beam  at  the  section.  The  direct 
shear,  however,  is  seldom  operative,  this  action  being  best  repre- 
sented by  a  punch  making  holes  in  a  plate  or  by  a  large  shear 
cutting  a  plate.  The  shearing  stress  which  breaks  a  beam  is  di- 
agonal tension  resulting  from  the  combined  action  of  the  horizontal 
and  vertical  shearing  stresses. 

The  direct  vertical  end  shear  is  equal  to  the  maximum  reaction. 
The  horizontal  shear  is  equal  in  amount  and  acts  along  the  neutral 
plane  where  the  fiber  stress  in  bending  changes  from  tension  to 
compression,  the  stress  being  in  reality  a  sliding  of  the  fibers  where 
they  have  no  bending  stress.  The  diagonal  tension  is  the  com- 
ponent of  these  two  actions.  (Fig.  60.)  Referring  again  to  the 
statement  that  the  area  of  the  shear  diagram  between  any  sec- 
tion and  the  nearest  support,  for  a  beam  resting  on  two,  or  more, 
supports,  equals  the  bending  moment  at  the  section,  the  unit 
shear  at  any  section  amounts  to 
_  V 
-jd 

in  which  s  =  unit  shear  in  pounds  per  square  inch, 

V  =  shear  at  the  section  in  pounds, 
MI  =  moment  in  inch  pounds  at  one  side  of  section, 
Mz  =  moment  in  inch  pounds  at  the  other  side  of  section, 
jd  =  moment  arm  in  inches. 

The  allowable  shearing  stress  in  steel  is 
10,000  Ibs.  per  square  inch.  After  obtain- 
ing the  size  of  beam  to  carry  a  certain  load, 
divide  the  maximum  reaction  by  the  web 
thickness  multiplied  by  the  depth  of  the 
beam.  This  will  give  the  shearing  stress  in 


Fig.  60  —  Relation  be-  pounds  per  square  inch.    If  it  exceeds  10,000 

tween  Moment  and      Ibs.  a  larger  beam  is  required.     In  the  steel 

Shear  handbooks  the  total  amount  of  shear  for 

which  a  beam  is  safe  is  given  in  the  tables  of  "  Properties  of 

Sections."     For  example,  in  taking  out  from  the  tables  a  beam 


INTERNAL  FORCES  77 

of  sufficient  size  to  carry  the  load,  the  total  amount  of  shear  the 
beam  is  good  for  should  be  equal  to  or  exceed  the  maximum  reac- 
tion. Thin  webs  act  like  long  slender  columns  and  may  fail  by 
crippling.  The  crippling  strength  of  the  beams  is  also  given  in  the 
tables  and  this  should  be  equal  to  or  exceed  the  maximum  reaction. 
Fig.  61  is  a  very  old  illustration  used  by  many  writers  to  explain 
shearing  action  in  a  beam.  Let  (a)  represent  a  beam  assumed 

to  be  composed  of  a 
number  of  planks  not 
(b)  fastened    together. 

Fig.  61  —  Illustration  of  Horizontal  Shear         When    loaded   the 

planks  bend  and  slide  on  each  other  as  shown.  This  sliding  action 
is  horizontal  shear,  which  is  zero  at  the  top  and  bottom  edges 
and  a  maximum  along  the  neutral  plane  where  the  tensile  stress 
changes  to  compressive  stress. 

Spike  the  planks  together  (6)  and  they  will  not  separate  when 
the  beam  bends  under  load.  The  sliding  stress  is  pure  shear 
on  the  spikes  connecting  the  planks.  The  spikes  act  by  bearing  on 
the  planks  into  which  they  are  driven,  and  in  this  manner  some 
tension  is  carried  from  outer  to  inner  planks.  The  resultant  force 
is  called  diagonal  tension. 

Imagine  a  beam  of  any  material  divided  into  a  great  number  of 
horizontal  layers.  Along  the  imaginary  joints  shear  exists  which 
is  resisted  by  the  tensile  strength  of  the  material.  In  beams  of 
steel  or  iron,  in  which  materials  the  tensile  strength  is  equal  in 
all  directions,  the  diagonal  tension  thus  developed  may  be  strong 
enough  to  tear  the  web  along  a  diagonal  line  extending  upward 
from  the  support.  The  web  must  be  thick  enough  to  resist  the 
diagonal  shear  or,  in  the  case  of  a  plate  girder,  be  strengthened 
by  stiffeners. 

Reinforced-concrete  beams  fail  similarly  in  diagonal  shear. 
This  may  be  resisted  by  making  the  stem  of  the  beam  thick  or, 
if  it  is  desirable  to  use  little  concrete  in  the  stem,  stirrups  are 
added  to  resist  diagonal  tension.  There  are  other  theories  which 
endeavor  to  account  for  the  diagonal  cracks  which  appear  some- 
tunes  in  reinforced-concrete  beams,  but  the  generally  accepted 
method  for  proportioning  stirrups  is  based  on  shear  expressed  as 
diagonal  tension,  and  since  the  desired  result  is  accomplished  and 
the  computations  are  readily  performed  this  method  it  is  believed 
will  persist. 


78  PRACTICAL  STRUCTURAL  DESIGN 

Wood  is  composed  of  actual  horizontal  fibers,  instead  of  the 
imaginary  fibers,  or  horizontal  planes,  considered  in  analyzing 
shear  in  beams  of  homogeneous  material.  If  wood  were  equally 
strong  in  all  directions,  a  shearing  failure  in  this  material  would 
also  be  indicated  by  the  appearance  of  diagonal  cracks. 

Shear  in  Wooden  Beams 

In  wooden  beams  the  dangerous  shear  acts  along  the  neutral 
plane  and  the  beam  may  split,  thus  by  shearing  action  being  con- 
verted into  two  shallow  beams,  which  will  then  break  by  bending, 
for  the  upper  half  must  carry  the  whole  load  and  the  lower  half 
carries  the  whole  load  when  the  upper  half  is  destroyed.  The 
strength  in  shear  of  wooden  beams  should  be  tested  by  the  follow- 
ing formula.  If  the  distributed  load  found  by  this  formula  is 
smaller  than  that  found  by  the  bending  formula,  increase  the  size 
of  the  beam. 


_ 

o 

in  which  W  =  the  load  the  beam  will  carry  without  failing  in  shear. 
6  =  breadth  in  inches, 
h  =  height  in  inches, 
s  =  shearing  stress  per  square  inch,  usually  one-tenth 

the  maximum  fiber  stress  in  bending. 
The  above  formula  is  derived  as  follows: 

W 
v  _W_      ,          _Z._JL_E        3        3TF 

~  Tai         "  jhb  ~  Ihb  ~  2  X  2hJb  ~  4hb 

,     ,       .  „. 

and,  therefore  W 

Modulus  of  Rupture 

The  modulus  of  rupture  is  a  measure  which  represents  a  com- 
bination of  all  the  forces  that  tend  to  break  a  beam  ;  i.e.,  the  com- 
bined action  of  tension  compression,  shear,  and  crippling.  It  was 
formerly  used  in  beam  design  to  obtain  the  breaking  load,  which 
was  divided  by  some  factor  of  safety  to  determine  the  safe  load. 
To-day  it  is  used  only  for  materials  in  which  it  is  difficult  to  sepa- 
rate the  different  stresses,  as,  for  example,  clay,  stone,  and  plain 
concrete.  The  moment  of  resistance,  using  an  allowable  safe 


INTERNAL  FORCES  79 

fiber  stress,  is  used  in  computations  for  beams  of  wood,  steel,  iron, 
and  reinforced  concrete. 

The  unit  moment  of  resistance  is  a  number  which  contains  all 
the  known  quantities  in  an  expression,  leaving  only  the  unknowns 
to  be  found.  For  example,  the  moment  of  resistance  of  a  wooden 
beam  in  which  we  can  use  a  maximum  fiber  stress  of  1200  Ibs.  per 

square  inch  is  1200  bh* 

Mr=— g— , 

and  by  dividing  the  fiber  stress  by  6  the  unit  moment  of  resistance 
equals  200,  from  which  we  get 

Mr  =  200  bhz  =  Rbh\ 

Some  men  use  R  for  wooden  beams,  but  where  the  divisor  is 
so  small  the  only  advantage  is  some  slight  simplification  of  the 
work,  provided  a  table  of  values  of  R  has  been  previously  com- 
puted for  the  woods  used.  In  reinforced-concrete  work  a  number 
of  factors  enter  into  the  formula  for  the  resisting  moment  and  the 
use  of  a  table,  or  of  a  diagram  which  is  really  a  graphical  table, 
for  all  possible  values  of  R  is  almost  indispensable  for  the  designer. 
Where  a  number  of  factors  enter  into  a  computation  it  is  easy  to 
forget  to  use  some. 

Deflection 

The  amount  of  deflection  when  a  beam  is  loaded  is  measured 
on  the  bottom  or  top  of  the  beam  for  convenience.  The  difference 
in  elevation  between  the  end  of  the  beam  and  the  middle  is  the 
deflection.  The  deflection  actually  used  in  computations  is  the  de- 
flection at  the  neutral  axis,  but  the  deflection  measured  on  the 
bottom  or  top,  which  for  obvious  reasons  is  more  readily  obtained 
than  the  deflection  of  the  neutral  plane,  is  close  enough  for  all 
practical  purposes. 

Deflection  in  beams  and  girders  used  in  buildings  is  important 
only  when  the  lower  side  carries  a  plastered  ceiling.  The  deflec- 
tion is  limited  to  a  maximum  of  one-three-hundred-and-sixtieth 
of  the  span  to  prevent  cracks  in  the  plaster.  A  greater  deflection 
is  not  unsightly  and  is  permissible  when  constant.  Wood  and  steel 
beams  straighten  when  the  load  is  relieved  and  deflect  when  the 
load  is  increased.  It  is  the  movement  that  causes  plaster  to  crack, 
so  this  must  be  limited.  For  beams  and  trusses  under  moving 
loads  the  deflection  must  be  limited  to  an  amount  which  will  not 
set  up  dangerous  vibrations,  but  with  this  the  ordinary  structural 


80  PRACTICAL  STRUCTURAL  DESIGN 

designer  seldom  has  to  deal,  it  being  part  of  the  work  involved 
in  the  design  of  bridges.  Deflection  also  affects  appearance  and 
a  camber  is  given  to  trusses  to  hide  deflection. 

A  beam  may  be  amply  strong  so  that  it  will  not  fail  by  bending, 
shearing,  or  crippling,  and  yet  the  deflection  may  be  so  great  that 
it  will  not  be  suitable  for  use  in  the  proposed  location.  The  amount 
of  deflection  must  then  be  found  and  if  it  exceeds  the  allowable 
deflection  a  deeper  beam  must  be  substituted.  When  using  steel 
it  is  often  possible  to  secure  a  deep  beam  which  will  weigh  less 
than  a  beam  of  less  depth  of  practically  equal  strength  in  bending 
and  shear.  For  timber,  experience  indicates  that  the  most  eco- 
nomical beam,  considering  the  two  factors  of  strength  and  stiffness, 
has  a  breadth  equal  to  two-thirds  or  three-fourths  the  depth. 

Deflection  Formulas 

Deflection  formulas  as  usually  presented  are  formidable  in 
appearance,  so  tables  are  given  in  the  steel  handbooks  which 
enable  the  deflection  in  inches  to  be  found  by  dividing  a  factor  in 
the  table  by  the  depth  of  the  rolled  section  in  inches. 

Similar  information  for  wooden  beams  was  not  so  readily 
obtainable  until  in  1913  the  Yellow  Pine  Manufacturers'  Associa- 
tion issued  a  book  entitled  "A  Manual  of  Standard  Wood  Con- 
struction," following  the  lines  laid  down  previously  by  the  steel 
manufacturers  in  their  handbooks.  Copies  of  this  book  may 
be  obtained  from  the  secretary  of  the  above  association  in 
St.  Louis,  Mo. 

The  "  Structural  Timber  Handbook,  for  Pacific  Coast  Woods  "  is 
issued  by  the  West  Coast  Lumbermen's  Association,  Seattle,  Wash., 
and  valuable  books  on  the  subject  of  wood  design  may  be  obtained 
from  the  National  Lumber  Manufacturers'  Association,  Chicago,  111. 

The  complicated  formulas  for  deflection  are  made  to  appear  as 
follows,  after  certain  substitutions  and  transformations  of  factors  : 


in  which  D  =  deflection  in  inches, 

/  =  allowable  maximum  fiber  stress  in  bending, 
L  =  length  in  feet, 
E  =  modulus  of  elasticity, 
h  =  height  of  beam  in  inches. 


INTERNAL  FORCES  81 

Assuming  common  values : 

For  steel,  /  =  16,000  Ibs.  per  square  inch,  D  =  ™-r* 

For  wood,  /  =  1300  Ibs.  per  square  inch,  D  =  TT~T* 

L2 
For  wood,  /  =  1000  Ibs.  per  square  inch,  D  =  TT~T' 

For  wood,  /  =  800  Ibs.  per  square  inch,  D  =  ^-r- 

See  page  98. 

Aids  to  Computation 

In  addition  to  the  handbooks  of  the  steel  companies  and  the 
"  Manual  of  Standard  Wood  Construction,"  designers  use  dia- 
grams and  slide  rules  to  lighten  their  work  on  simple  problems. 
The  following  are  suggested  in  this  connection: 

The  Wager  timber  scale  for  computing  the  strength  of  wooden 
beams,  $1. 

The  Merritt  beam  scale  for  computing  the  strength  of  steel 
beams,  $1. 

Des  Moines  Bridge  &  Iron  Company's  calculator  for  steel 
beams,  channels,  angles,  and  tees,  25  cents. 

The  two  first  mentioned  are  made  of  heavy  paper  and  the 
third  is  of  celluloid.  The  writer  has  used  them  daily  in  his  work 
for  some  years. 

The  most  complete  rule  for  this  work  is  one  designed  by  Benja- 
min Winslow.  It  enables  one  to  design  with  any  fiber  stress,  any 
span,  any  spacing,  any  system  of  loading,  etc.  The  rule  is  made  of 
German  silver  and  costs  $10.  The  size  is  3^  ins.  x  10|  ins.  X  16  ins. 
Taking  the  place  as  it  does  of  all  pocket  books,  tables  and  diagrams, 
the  writer  feels  that  to  omit  recommending  it  to  structural  drafts- 
men and  designers  would  be  a  neglect  on  his  part  of  a  plain  duty. 
Mr.  Winslow  has  also  placed  on  the  market  a  similar  slide  rule 
for  reinforced-concrete  design. 

Slide  rules  of  the  Mannheim  type  are  used  to-day  by  all  'engi- 
neers, but  a  recent  improvement  is  known  as  the  Phillips  slide 
rule.  This  rule  enables  one  to  multiply  three  factors  at  one 
setting  and  the  arrangement  of  the  graduations  wonderfully  in- 
creases the  value  of  the  slide  rule  for  all  purposes.  This  new  rule 
sells  for  $5. 

Example.  —  Determine  the  size  of  a  wooden  beam    using  a 


82  PRACTICAL  STRUCTURAL  DESIGN 

maximum  fiber  stress  of  1000  Ibs.  per  square  inch  to  carry  a 
uniformly  distributed  load  of  6000  Ibs.  on  a  span  of  14  ft. 

Answer.  —  Assume  the  beam  to  weigh  20  Ibs.  per  linear  foot 
=  14  x  20  =  280  Ibs.    The  total  load  =  6280  Ibs. 

M  =  628°  X014  X  12  =  1.5  x  6280  x  14  =  131,880  in.  Ibs. 

o 

Assume  a  depth  of  12  ins.,  which  will  give  a  beam  11.5  ins.  the 
usual  depth  of  a  commercial  size  12-in.  beam. 

Mr  =  ^-  =  167  bh2  =  131,880  in.  Ibs. 
o 

Mr  131,880 


This  calls  for  a  commercial  size  beam  7  ins.  X  12  ins.,  the  actual 
size  of  which  will  probably  be  about  6.5  ins.  x  11.5  ins.  The  weight 
per  linear  foot,  assuming  wood  to  weigh  35  Ibs.  per  cubic  foot,  will 

be  6'5  X  \\'f  X  35  =  18.2  Ibs.     This  is  so  close  to  the  weight 

J.44 

assumed  that  we  will  let  it  stand. 
Investigate  for  shear. 

4x  6.5  X  11.5  X  100 


O  O 

The  load  of  6280  Ibs.  is  therefore  safe. 
The  deflection  is  to  be  kept  below  -jg-^  of  the  span 

12x14 


360 


=  0.466  in. 


14x14  L 

=  0.38  in.  =  7-77: • 


44  X  11.5  443 

Find  the  span  on  which  the  safe  bending  load  is  equal  to  the  safe 
shearing  load. 

_8jtf_      143,500 
"  12  W  ~  1.5  X  9967 

This  beam  cannot  be  safely  loaded  with  more  than  9967  Ibs. 
on  any  span  of  less  than  9  ft.  8  ins.  no  matter  what  the  safe  load 
in  bending  may  be.  (The  moment  used  here  is  the  actual  resisting 
moment  of  the  beam  which  had  to  be  selected  to  carry  the  load, 
that  is,  M  =  167  x  6.5  X  11.52  =  143,500  in.  Ibs.,  the  actual  bend- 


INTERNAL  FORCES  83 

ing  moment  as  we  have  seen  being  131,880  in.  Ibs.  It  is  cheaper  to 
use  a  commercial  beam  with  a  resisting  moment  larger  than  the 
bending  moment  than  to  trim  the  beam  down  to  the  theoretically 
exact  size.  This  happens  with  rolled  steel  beams  also.  When  a 
built-up  plate  girder  or  a  latticed  girder  (truss)  is  used  the  differ- 
ence between  the  bending  moment  and  resisting  moment  can  be 
cut  to  a  smaller  amount.  Reinforced  concrete  is  a  material  which 
permits  of  closer  designing  than  rolled  shapes,  hence  the  differ- 
ences in  design  shown  by  equally  competent  designers  tackling 
the  same  problem  when  using  reinforced  concrete. 

A  formula  to  find  the  limiting  span  when  bending  and  shear 
are  considered  is  developed  as  follows  for  wood :  M  in  inch  pounds. 
8M  M  3M       M 


L  = 


12W      1.5  x  4bhs      Qbhs      2bhs 


Find  the  deflection  on  the  li  mi  ting  span. 

Q  67  v  1  2 
The  allowable  deflection  =  =  0.323  in. 


The  actual  deflection  =  -       =  '      =  0.184  in. 

44/i        44  x  11.5 

Find  the  allowable  safe  uniformly  distributed  load  the  beam 
will  carry  on  a  span  of  20  ft. 

„,      8  M       143,500 

^  =  i2L  =  L5t2( 

12  v  20 
Allowable  deflection  =       *       =  0.667  in. 

ouU 

Actual  deflection  =  ,2.°  X^°K  =  0.79  in. 
44  X  11.5 

The  deflection  is  too  great  if  the  lower  side  of  the  beam  is  to  be 
plastered,  or  the  beam  is  to  carry  a  plastered  ceiling. 

NOTE.  —  When  a  wooden  beam  has  a  depth  in  inches  less  than 
two-thirds  the  span  in  feet  the  deflection  is  apt  to  cause  plaster  to 
crack.  Try  a  beam  14  ins.  deep,  the  actual  depth  being  13.5  ins. 
,  _  Mr  143,500 

"  ~  167  x  13.52  * 


Try  a  commercial  6  ins.  X  14  ins.  =  5.5  ins.  X  13.5  ins. 

Mr  =  167  x  5.5  x  13.52  =  167,500  in.  Ibs. 
This  beam  is  seen  to  be  excessively  strong,  but  a  beam  4.5  ins.  x  13.5 


84  PRACTICAL  STRUCTURAL  DESIGN 

ins.  would  have  a  resisting  moment  of  only  137,000  ins.  Ibs.    Allow- 
able deflection  =  0.667  ins.     Actual  deflection 

20  X  20          AA__  . 
=  TA  -  TIT-?  =  0.6675  ms. 
44  x  13.5 

The  deflection  in  the  formulas  presented  is  dependent  upon 
the  stress,  so  the  deflection  found  is  that  produced  when  the  beam 
is  fully  stressed,  that  is,  when  the  full  resisting  moment  of  167,500 
in.  Ibs.  is  developed.  Under  the  load  found  for  the  20-ft.  span  the 
moment  is  only  143,500  in.  Ibs.,  so  the  deflection  will  be  less  than 
that  given. 

This  case  may  be  dealt  with  as  follows  if  it  is  desired  to  find  the 
actual  deflection.  The  divisor  for  the  span  squared  is  41  for  a 
fiber  stress  of  1300  Ibs.  per  square  inch,  44  for  a  fiber  stress  of  1000 
Ibs.  per  square  inch,  and  46  for  a  fiber  stress  of  800  Ibs.  per  square 
inch.  The  divisor  is  seen  to  alter  by  1  for  each  100  Ibs.  change  in 
unit  fiber  stress.  Find  the  maximum  fiber  stress  for  the  bending 
moment  developed  and  then  applying  the  proper  divisor  ascertain 
the  actual  deflection. 

143,500      6  x  143,500 
/  -  -^-  =   5  5  x  13  .2   =  859.2  Ibs.  per  square  inch. 

6 
The  divisor  for  all  practical  purposes  is  45.6. 

L2  20  x  20 

D  =  -  45.6  X  13.5  =  °'65  mS" 


The  load  this  beam  can  carry  on  a  20-ft.  span  with  a  deflection 
equal  to  0.6675  in.  is 


All  the  computations  have  been  made  with  a  slide  rule,  so  in 
some  cases  the  terminal  figures  in  the  results  may  differ  slightly 
from  those  found  by  arithmetical  computations,  but  when  deal- 
ing with  large  quantities  small  differences  in  the  units  place  make 
no  material  difference  in  results. 

The  calculations  for  deflection  in  wooden  beams  can  never 
give  exact  results,  for  woods  vary  in  texture  throughout  and  the 
amount  of  moisture  and  seasoning  also  act  to  increase  or  decrease 
deflection. 


CHAPTER  III 
Problems  in  Design  of  Beams 

THE  two  standard  steel  handbooks  are  the  "  Carnegie  Pocket 
Companion  "  and  the   "  Cambria  Steel  Manual."     The 
designer  should  have  one  *or  both  of  these  books.    The 
Bethlehem  Steel  Company  issues  a  handbook  which  the  designer 
should  also  possess,  owing  to  the  differences  in  shape  and  carrying 
capacity  of  the  Bethlehem  and  standard  beams. 

The  Cambria  and  Carnegie  handbooks  contain  a  great  deal  of 
text  book  matter  and  are  very  useful  to  students  and  to  men  who 
wish  occasionally  to  refresh  their  memories  on  points  of  design. 
They  contain  the  usual  tables  indispensable  to  structural  de- 
signers. The  handbook  of  the  Lackawanna  Steel  Company  and 
that  issued  by  Jones  &  Laughlin  contain  the  indispensable  tables 
and  some  memory  aiding  text,  but  not  as  much  as  the  first  books 
mentioned. 

For  a  uniformly  distributed  load  the  size  of  a  beam  is  easily 
obtained.  Tables  give  the  uniformly  distributed  loads  in  pounds 
for  all  spans,  varying  by  single  feet  which  the  different  beams  can 
carry.  By  reducing  concentrated  loads  to  their  equivalents  in 
uniformly  distributed  loads  these  tables  may  be  used  for  any 
system  of  loading  without  first  ascertaining  the  bending  moment. 

When  concentrated  loads  are  dealt  with  as  such  and  the  bend- 
ing moments  are  found,  the  proper  size  beam  may  be  found  by 
looking  up  the  bending  moment  in  foot  pounds,  opposite  which, 
on  the  same  line,  is  found  the  size  and  weight  of  the  beam.  Beams 
must  be  secured  (stayed)  laterally  to  prevent  side  bending ;  other- 
wise the  carrying  capacity  is  less  than  that  given  in  the  tables. 

The  Carnegie  book  formerly  gave  a  factor  of  strength,  C,  to 
use  when  the  bending  moment  was  used.  It  is  designated  as  C 
in  the  Bethlehem  book  and  as  F  in  the  Cambria  book.  In  the  1913 
edition  of  Carnegie  this  factor  is  not  given,  the  bending  moment 
in  foot  pounds  being  shown  on  the  page  containing  the  other 
properties  of  beams. 

85 


86  PRACTICAL  STRUCTURAL  DESIGN 

The  factor  of  strength  is  obtained  as  follows: 

The  fiber  stress  is  in  pounds  per  square  inch  and  the  section 
modulus  is  in  square  inches,  therefore  the  resisting  moment  is  in 
inch  pounds,  or  12  times  the  bending  moment  in  foot  pounds. 
Two-thirds  the  moment  in  inch  pounds  is  equal  to  8  times  the 
bending  moment  in  foot  pounds;  which,  in  turn  is  equal  to  the 
total  uniform  load  in  pounds  times  the  span  in  feet,  for  a  freely 
supported  beam. 

Let  S  =  section  modulus  in  inches. 

/  =  maximum  fiber  stress  in  Ib.  per  sq.  in. 
Then  C  =  F  =  $fS. 

Let  M  =  bending  moment  in  foot  pounds. 

Then  C  =  F  =  SM. 

Having  computed  the  bending  moment  in  foot  pounds,  multiply 
by  8  and  in  the  table  of  properties  of  beams  look  for  this  value, 
or  the  nearest  higher  value,  of  F  (or  of  C)  in  the  Cambria  or  Beth- 
lehem book.  Following  the  line  to  the  right,  the  beam  is  found 
which  has  this  factor  of  strength.  Each  of  the  books  mentioned 
contains  a  separate  table  of  bending  moments  in  foot  pounds  for 
each  beam,  so  the  designer  has  his  choice  of  methods  to  use  in 
obtaining  a  beam  size  when  he  has  the  bending  moment  instead 
of  the  uniformly  distributed  load. 

Example.  —  A  beam  carrying  several  concentrated  loads  must 
resist  a  bending  moment  of  46,680  ft.  Ibs.  What  is  the  best  size 
and  weight  of  beam  to  use? 

Carnegie  (1913  edition) :  On  page  184  it  is  shown  that  the 
resisting  moment  of  a  12-in.  I-beam  weighing  31.5  Ibs.  per  Hn.  ft. 
=  47,960  Ibs.,  so  this  beam  will  be  used. 

Page  182  contains  a  description  of  all  the  factors  shown  on 
page  184,  relating  to  the  properties  of  beams.  The  student  is  now 
prepared  to  study  pages  133,  140,  141,  164,  167  to  171  inclusive, 
176  to  182  inclusive. 

Cambria  (1913  edition) :  On  page  118  it  is  shown  that  a  12-in. 
I-beam  weighing  31.5  Ibs.  per  lin.  ft.  has  a  resisting  moment  of 
48,000  ft.  Ibs. 

The  following  pages  should  be  studied  by  the  student,  76,  77, 
80  to  89  inclusive,  142  to  147  inclusive,  158  to  163  inclusive. 

"Lackawanna  Hand  Book"  (1915  edition):  This  book  does 
not  contain  a  table  of  bending  moments  for  standard  beams  so 


PROBLEMS  IN  DESIGN  OF  BEAMS  87 

the  bending  moment  in  foot  pounds  must  be  multiplied  by  8  and 
the  tables  on  pages  164  to  167  consulted.  The  size  of  beam  is 
given  in  Col.  12  on  page  167  and  we  find  that  a  12-in.  I-beam, 
weighing  31.5  Ibs.  per  lin.  ft.,  will  be  required.  On  the  pages 
mentioned  is  a  column  containing  distances  center  to  center  of 
beams  required  to  make  the  radii  of  gyration  equal;  a  very  useful 
table  to  use  when  designing  columns. 

In  this  book  the  student  should  read  carefully  pages  144  to 
163  inclusive. 

Jones  &  Laughlin,  "Standard  Steel  Construction"  (1916 
edition :  This  book  does  not  contain  a  table  of  bending  moments 
for  I-beams,  neither  does  it  contain  a  table  of  factors,  C.  or  F. 
Our  problem  is  solved  as  follows;  Since  the  moment  divided  by 
the  fiber  stress  equals  the  Section  Modulus,  divide  the  bending 
moment  in  foot  pounds  by  the  fiber  stress,  16,000  Ibs.,  and  this 
gives  the  section  modulus  in  feet.  Multiply  by  12  to  obtain 
the  section  modulus  in  inches.  Look  up  this  value  on  pages 
105-106.  Proceeding  in  this  fashion  we  get  (46,680  -r-  16,000) 
X  12  =  35  in.  =  S.  On  page  106  the  nearest  value  is  36,  cor- 
responding to  a  12-in.  I-beam  weighing  31.5  Ibs.  per  lineal  foot. 

The  student  should  now  become  familiar  with  pages  95  to  175 
inclusive,  and  with  page  243  in  this  book. 

Bethlehem  (1911  edition):  On  page  38  a  9-in.  girder-beam 
weighing  38  Ibs.  per  lin.  ft.  has  a  resisting  moment  of  50,630  Ibs. 
On  page  39  a  12-in.  Bethlehem  I-beam  weighing  28.5  Ibs.  per  lin.  ft. 
has  a  resisting  moment  of  48,050  ft.  Ibs. 

To  understand  why  the  Bethlehem  beams  are  stronger  than 
standard  I-beams  of  equal  depth,  read  pages  3  to  9  inclusive.  Then 
study  pages  30,  31,  56,  66,  68,  99  to  103  inclusive. 

In  studying  the  pages  mentioned  the  student  should  work 
examples  in  order  to  become  familiar  with  the  use  of  the  tables. 
The  tables  of  deflection  factors  should  be  thoroughly  understood, 
which  is  not  a  difficult  matter  if  the  remarks  on  deflection  in  this 
chapter  have  been  given  proper  attention. 

After  thoroughly  mastering  the  subject  matter  on  the  pages 
enumerated  the  student  should  study  pages  283  to  292  inclusive 
in  Carnegie ;  56  to  71  inclusive  in  Cambria ;  104  to  107  in  Beth- 
lehem. The  pages  mentioned  in  each  book  cover  the  same  sub- 
jects, so  it  is  not  necessary  to  use  the  three  books,  one  giving  all 
that  is  necessary.  Should  the  student,  however,  possess  the  three 


88  PRACTICAL  STRUCTURAL  DESIGN 

books,  it  will  be  well  to  study  the  subjects  thoroughly  in  one 
and  then  become  familiar  with  the  similar  matter  presented  in 
the  others. 

Only  rolled  shapes  have  been  considered  so  far.  Compound 
shapes,  i.e.  plate  girders  and  trusses,  will  be  taken  up  later. 

Practical  Problems  in  Design 

1.  Find  the  resisting  moment  of  flooring  f  in.  thick;    £  in. 
thick ;  1|  in.  thick ;  If  ins.  thick. 

Answer.  —  f  in.  =  0.625  in. ;  £  in.  =  0.875  in. ;  If  ins.  =  1.125 
ins. ;  If  ins.  =  1.75  ins.  The  width,will  be  taken  as  12  ins.,  as  floor 
loads  are  generally  given  in  pounds  per  square  foot.  The  flooring 
is  white  pine  having  a  fiber  stress  of  800  Ibs.  per  square  inch. 
[In  all  problems  it  is  understood  that  by  fiber  stress  is  meant 
the  maximum  (skin)  stress.]  The  unit  moment  of  resistance 
=  800  -r-  6  =  133.33. 

Mr  =  133.33  X  12  x  0.6252  =    625  in.  Ibs. 

Mr  =  133.33  x  12  x  0.8752  =  1225  in.  Ibs. 

Mr  =  133.33  x  12  x  1.1252  =  2025  in.  Ibs. 

Mr  =  133.33  X  12  x  1.752    =  4900  in.  Ibs. 

2.  What  is  the  greatest  spacing  permissible  between  joists  if 
the  deflection  is  to  be  limited  the  usual  amount? 

Flooring  comes  in  long  pieces  and  thus,  extending  over  a  number 
of  supports,  to  each  of  which  it  is  nailed,  the  thickness  can  be  equal 
in  inches  to  one-half  the  span  in  feet.  This  gives  a  maximum  span 
for  the  |-in.  of  2  x  0.625  =  1.25  ft.  (15  ins.) ;  |-in.,  2  x  0.875  =  1.75 
ft.  (21  ins.) ;  H-ins.,  2  x  1.125  =  2.25  ft.  (27  ins.) ;  If-ins.,  2  x  1.75 
=  3.5  ft.  (42  ins.). 

Floors  generally  have  greater  stiffness  than  is  here  shown 
because  of  the  tongue  and  groove  along  the  edges,  but  this  is  fre- 
quently nullified  by  the  fact  that  the  loads  brought  on  floors  are 
more  often  concentrated  than  uniformly  distributed.  The  above 
rule  for  deflection  is  arbitrary,  and  if  the  spans  mentioned  are 
actually  used  it  will  be  well  to  check  the  deflection  by  a  proper 
formula.  Refer  to  the  table  of  relative  strength  and  stiffness  of 
beams.  The  deflection  formula  gives  deflection  for  uniform  loads 
on  beams  resting  freely  on  two  end  supports.  First  find  the  de- 
flection by  the  formula  and  multiply  it  by  the  constant  found  in 
the  column  of  relative  deflections,  opposite  the  condition  of  loading 
to  which  the  case  under  consideration  may  apply. 


PROBLEMS  IN  DESIGN  OF  BEAMS  89 

3.  Neglecting  deflection,  what  is  the  greatest  permissible  spacing 
of  joists  for  the  following  loads  per  square  foot  (including  the  weight 
of  the  flooring)  :  42  Ibs.  ;  781bs.;  103  Ibs.;  129  Ibs.? 

Flooring  extends  over  several  supports,  so  we  may  assume  a 
condition  of  restraint  and  use  the  formula 

,,      wU  .    .     . 

M  =  —  2~,  in  foot  pounds. 

The  load  is  given  in  pounds  per  square  foot,  so  the  span  should 
be  in  feet.  The  formula  then  becomes 


,  ,  .     ., 

M  =  in.  Ibs. 

which  reduces  to  M  =  wl?  in.  Ibs.  % 


ci-    -i    i     f      Tkr  .     „ 

Similarly,  for  M  =  —  5  —  in.  Ibs. 

o 

we  obtain  M  =  l.5wL?  in.  Ibs. 

Another  condition  sometimes  met  with  in  wood  and  steel  design 
and  frequently  used  in  reinforced  concrete  design  is  a  partially 
restrained  condition  in  which  the  beam  rests  freely  on  one  end 
support  and  is  fully  restrained  at  the  other  support.  For  this  con- 
dition the  coefficient  is  10  and 


M  =  —       in.  Ibs.,  or  M  =  l.2wL\ 

Using  the  expression  M  =  wl?,  the  spans  for  the  various  floor 
thicknesses  are  found  as  follows: 

M  4/M 

L2  =  —  .  or  L  =  V/  — 
w  V  w 

Using  the  resisting  moments  in  inch  pounds  obtained  for  each 
thickness, 


f-in.  flooring:  L  =  =  3.85  ft. 


The  rest  of  the  examples  are  left  to  the  student  as  a  useful 
exercise. 

4.  A  floor  is  constructed  of  2-in.  (1.75-in.)  planking  laid  over 
beams  spaced  4  ft.  6  ins.  center  to  center,  the  span  of  the  beam  from 
wall  to  girder  being  18  ft.  Find  size  of  beam  when  the  total  load 


90  PRACTICAL  STRUCTURAL  DESIGN 

per  square  foot,  including  weight  of  beam  and  floor,  is  132  Ibs.  per 
square  foot.  Material  yellow  pine  with  an  allowable  fiber  stress 
of  1300  Ibs.  per  square  inch.  Deflection  ignored. 

Answer.  —  The  total  load  on  the  panel  is  132  x  4.5  X  18 
=  10,700  Ibs. 

M  =  1.5  X  10,700  x  18  =  288,900  in.  Ibs. 

Try  an  8-in.  x  14-in.  beam  (7.5  ins.  x  13.5  ins.) 

, ,       1300  x  7.5  x  13.52      _n_  .  ._  .     ., 

Mr  = ^ =  296,156  in.  Ibs. 

6 

Try  for  shear 

w_«*_  4X7.5X13.5X130 

O  O 

W.  N.  Twelvetrees,  a  British  engineer,  developed  the  following 
method  for  designing  a  beam  in  which  the  breadth  is  to  be  some 
definite  proportion  of  the  depth. 

Let  n  =  r,  then  6  =  — 
V  n 


To  design 

soft 

h 

=  2 

h 

1 

=  2 

b 

2h 

=  y 

_  A_ 

1 

0.67 

=  1.5 

07, 

h 

1 

b 

=  T       1 

075 

=  1.33 

M  =  RW 

h 
=  R  x- 

n 

X/i2 

n 

Applying  the  method  to  the  example  under  consideration: 

Let  R  =  t  =  ^  =  217. 
6         6 

First.  —  Design  so  the  breadth  equals  one-half  the  depth. 


Use  commercial  size  7.5  ins.  X  14.5  ins. 

Second.  —  Design  so  the  breadth  equals  two-thirds  the  depth. 


PROBLEMS  IN  DESIGN  OF  BEAMS  91 

Substituting  in  the  formula  n  =  1.5,  find  h  =  12.6  ins. 

2  x  12.6 

6  =  — 5 =  8.4  ins. 

o 

Use  a  commercial  size  beam  9  ins.  x  13  ins. 

Third.  —  Design  so  the  breadth  equals  three-quarters  the  depth. 

Substituting  in  the  formula  n  =  1.33,  find  h  =  12.08  ins. 

,      3  x  12.08 

6  = -A =  9.06  ins. 

4 

Use  commercial  size  beam  9.5  ins.  x  12.5  ins. 

The  student  will  have  noticed  that  in  all  cases  the  exact  size 
computed  cannot  be  used  and  it  is  necessary  to  take  a  commercial 
size  enough  larger  so  the  loss  in  dimensions  through  cutting  will 
give  a  beam  the  size  of  the  computed  beam,  or  slightly  larger. 
Small  beams  will  run  from  j  in.  to  f  in.  smaller  than  nominal  size, 
but  beams  of  the  size  here  considered  will  seldom  run  less  than 
|  in.  smaller  in  each  dimension  than  the  nominal  size,  and  if  the 
superintendent  of  construction  is  not  careful  the  loss  will  be 
even  greater.  The  writer  is  acquainted  with  designers  who  use 
the  nominal  size  always  in  their  designs,  assuming  that  the  maxi- 
mum fiber  stress  allowed  is  really  less  than  the  wood  can  stand. 
It  is  not  good  practice. 

Assuming  that  the  fiber  stresses  are  based  on  the  use  of  wood 
freely  exposed  to  weather,  then  the  following  increases  in  fiber 
stress  are  allowable  for  long-leaf  yellow  pine: 

Class  A  (moisture  contents,  18  per  cent). — Structures  freely 
exposed  to  the  weather,  such  as  railway  trestles,  uncovered  bridges, 
etc.,  let  allowable  stress  equal  1  x  /. 

Class  B  (moisture  contents,  15  per  cent).  —  Structures  under 
roof  but  without  side  shelter,  freely  exposed  to  outside  air,  but 
protected  from  rain,  such  as  roof  trusses  of  open  shops  and  sheds, 
covered  bridges  over  stream,  etc.,  let  allowable  stress  equal  1.15  x  /. 

Class  C  (moisture  contents,  12  per  cent).  —  Structures  in  build- 
ings unheated,  but  more  or  less  protected  from  outside  air,  such 
as  roof  trusses  of  barns,  inclosed  shops  and  sheds,  etc.,  let  allow- 
able stress  equal  1.4  x/. 

Class  D  (moisture  contents,  10  per  cent).  —  Structures  in 
buildings  at  all  times  protected  from  the  outside  air,  heated  in 
the  winter,  such  as  roof  trusses  in  houses,  halls,  churches,  etc., 
let  allowable  stress  equal  1.55  x/. 


92 


PRACTICAL  STRUCTURAL  DESIGN 


For  all  woods  other  than  long-leaf  yellow  pine  the  increases  to 
be  one-half  those  given.  The  shearing  stress,  however,  cannot 
exceed  one-tenth  the  fiber  stress  used  for  Class  A  structures. 

Building  ordinances  in  American  cities  do  not  recognize  any 
difference  in  allowable  stresses  dependent  on  the  moisture  contents, 
so  the  fiber  stresses  permitted  in  cities  apply  to  all  structures. 
It  would  be  better  if  the  city  ordinance  requirements  were  based 
on  Class  D  structures  with  proportionate  decrease  for  structures 
in  other  classes. 

The  following  table  gives  the  allowable  fiber  stresses  for  wood 
in  the  city  of  Chicago  (1916).  Each  designer  should  use  the  stresses 
permitted  in  the  largest  city  nearest  to  the  place  where  the  building 
is  to  be  erected. 


Maximum 

•p:  Up- 

Name  of  the  Wood 

Bending 
Stress 
and  Ten- 

Com- 
pression 
with 
Grain 

Com- 
pression 
across 
Grain 

Shear 
with 
Grain 

sion  with 

Grain 

Douglas  fir  and  long-leaf  yellow  pine 

1300 

1100 

250 

130 

Oak  

1200 

900 

500 

200 

Short-leaf  yellow  pine  

1000 

800 

250 

120 

Norway  pine  and  white  pine 

800 

700 

200 

80 

Hemlock  

600 

500 

150 

60 

The  first  column  gives  the  name  of  the  wood.  The  second  column 
gives  the  maximum  bending  fiber  stress  and  this  is  the  maximum 
stress  allowed  if  the  wood  is  to  be  used  as  a  tie  in  straight  tension 
—  something  rarely  possible  because  of  the  difficulty  in  making 
proper  connections  so  the  nails,  screws  or  bolts  will  properly  trans- 
mit the  entire  pull  on  the  piece. 

The  third  column  gives  the  compressive  stress  per  square  inch 
on  wood  posts  having  a  least  breadth  one-fifteenth  the  length. 
For  lengths  greater  than  fifteen  times  the  least  dimension,  the 
compressive  stress  must  be  reduced,  by  a  formula  given  in  the 
ordinance,  long  slender  pieces  bending  under  load  and  causing 
additional  strain  on  the  concave  side. 

The  fourth  column  gives  the  allowable  bearing  stress  per  square 
inch  on  the  under  side  of  a  beam  on  the  supports.  The  reaction 


PROBLEMS  IN  DESIGN  OF  BEAMS  93 

is  to  be  divided  by  the  stress  given  in  this  column  in  order  to  obtain 
the  number  of  square  inches  bearing  surface.  The  student  should 
pay  attention  to  this  column,  for  it  explains  the  reason  why  steel 
and  iron  post  caps  are  used  instead  of  the  old-fashioned  wooden 
bolsters.  If  the  load  on  a  column  is  carried  straight  down  on  the 
ends  of  fibers,  the  full  bearing  capacity  of  the  wood  can  be  utilized. 
When  a  bolster  is  set  between  the  foot  of  a  post  on  one  floor 
and  the  top  of  the  post  on  the  floor  below,  the  compression  across 
the  grain  of  the  wood  in  the  bolster  governs  the  carrying  capacity 
of  the  post,  or  the  bolster  will  crush. 

The  fifth  column  gives  the  allowable  shearing  stress  with  the 
grain,  the  use  of  this  column  having  been  explained  in  the  examples 
when  a  test  was  made  of  the  weight-carrying  capacity  of  a  beam 
BO  it  would  not  fail  in  shear. 

There  is  a  shear  parallel  with  the  grain  and  if  through  some 
unavoidable  circumstance  it  ever  becomes  necessary  to  design  so 
a  wide  beam  overhangs  the  sides  of  a  support  this  shear  will  act. 
It  should  not  exceed  the  safe  shear  with  the  grain. 

There  is  a  shear  across  the  grain,  that  is,  a  tendency  for  the 
beam  to  be  cut  at  the  edge  of  the  support.  Provided  the  allow- 
able compression  across  the  grain  is  not  exceeded,  i.e.,  sufficient 
bearing  surface  is  provided,  the  effect  of  this  shear  is  negligible. 

The  use  of  hangers  and  stirrups  is  common  to-day.  They  save 
head  room  but  increase  the  insurance  rate,  for  the  reason  that  metal 
is  affected  by  intense  heat.  A  large  piece  of  timber  will  char  on 
the  surface  and  must  be  exposed  to  an  intense  flame  for  a  long 
time  before  it  begins  to  burn.  The  heat  that  will  merely  char  a 
timber  and  do  it  little  harm  will  heat  wrought  iron  and  steel  to 
such  an  extent  that  the  stirrup  will  be  weakened  and  permit  the 
suspended  beam  to  drop.  A  study  of  a  bending  moment  curve 
shows  that  at  the  bearing  end  of  a  beam  there  is  practically  no 
moment,  so  the  area  of  a  beam  may  be  reduced  nearly  one-half 
at  the  supports  without  impairing  the  bearing  capacity.  If  the 
strength  of  a  stirrup  is  reduced  one-half  by  fire  the  beam  may  drop. 

Many  types  of  stirrups  are  on  the  market,-  and  before  adopting 
anything  other  than  a  plain  bent  strap  of  steel  or  wrought  iron 
the  designer  should  require  the  manufacturers  to  furnish  records 
of  tests  on  the  stirrups  they  propose  to  supply. 

To  design  a  stirrup:  First  obtain  the  area  required  for  bearing, 
then  the  thickness  to  prevent  straightening  at  the  edge  of  the  sup- 


94  PRACTICAL  STRUCTURAL  DESIGN 

port,  then  check  to  see  that  the  area  of  the  vertical  legs  is  sufficient 
in  tensile  strength  to  carry  the  load.  This  last  item  is  generally 
taken  care  of  when  the  other  conditions  are  satisfied. 

5.  Design  a  strap  hanger,  or  stirrup,  for  the  8-in.  x  14-in.  beam 
in  the  last  example. 

Answer.  —  The  total  load  =  10,700  Ibs.  which  gives  a  reaction 
=  5350  Ibs.  The  allowable  compression  across  the  grain  =  250  Ibs. 

eo-n 

per  square  inch,  so  the  bearing  area  in  the  stirrup  =  -  =  21.4 
sq.  ins.  The  width  of  the  beam  is  8  ins.,  therefore  the  width  of  the 
strap  under  the  end  of  the  beam  =  — ^-  =  2.66  ins.  Make  the  strap 

o 

2.75  ins.  wide,  a  stock  width.    Allowing  a  value  of  10,000  Ibs.  per 


Fig.  62  —  Various  Styles  of  Stirrups 
square  inch  tension  for  wrought  iron  the  required  area  of  the  two 

corn 

legs  =    Q         =  0.535  sq.  ins.  or  0.2675  sq.  ins.  for  each  leg.    The 

thickness  of  metal  required  =    '          =  0.097  in.  (practically  No. 

Z.to 

10  gauge).    Allowing  a  fiber  stress  of  14,000  Ibs.  per  square  inch 

corn 

for  steel,  the  required  area  in  the  two  legs  =       ^^  =  0.382  sq. 


in.,  or  0.191  sq.  in.  for  each  leg.     The  thickness  of  metal  required 

=    0  -g  =  0.0695  in.  (practically  No.  13  gauge).    Each  leg  must 
^./o 

rest  on  top  of  the  girder  with  a  length  of  not  less  than  4  ins. 

This  is  thin  metal  and  will  surely  straighten  under  the  load, 
besides  which  it  does  not  offer  enough  body  to  resist  corrosion. 
Use  a  minimum  thickness  of  f  in.  The  stirrup  shown  in  Fig.  62 
is  double  and  the  weight  of  the  beam  on  either  side  tends  to  bal- 
ance the  weight  of  the  beam  on  the  opposite  side  of  the  girder. 
A  stirrup  2|  ins.  wide  of  j-in.  metal  will  therefore  be  all  right  and 
may  be  wrought  iron  or  steel.  A  couple  of  holes  drilled  through 


PROBLEMS  IN  DESIGN  OF  BEAMS  95 

the  top  for  lag  screws  or  spikes  will  take  care  of  unequal  loading 
on  beams. 

When  a  half  stirrup  is  used  it  must  be  investigated  for  bending 
at  the  bearing  edge.    Assume  a  bearing  length  of  4  ins. 
4 


This  computation  considers  the  legs  as  cantilever  beams  uniformly 
loaded. 

The  thickness  of  the  metal  is  f  in.  (0.375  in.)  and  the  width  is 
to  be  found.    For  wrought  iron  with  a  fiber  stress  of  10,000  Ibs.  per 

10  000 
square  inch,  R  =  —  —  =  1667.     For  steel  with  a  fiber  stress  of 


. 

14,000  Ibs.  per  square  inch,  R  =  -  =  2333. 

*_M_  5350 

"  Rh*  ~  1667  x  0.3752  " 
of  which  each  leg  will  be  one-half,  or  11.4  ins.  for  wrought  iron. 

COKQ 

For  steel,  b  =  ^^  —  =  16.3  ins.,  of  which  each  leg  will  be 

Zooo  X  U.oiO 

one-half,  or  8.15  ins. 

The  reason  for  the  low  stresses  used  is  due  to  the  blacksmith 
work  required  to  bend  the  metal  to  the  required  shape,  the  heating 
annealing  the  metal  and  restoring  disturbed  molecules  to  a  normal 
condition.  Cold  working  has  a  contrary  effect  —  within  limits  — 
but  may  crack  the  metal,  thus  nullifying  the  effect  of  the  strain 
which  sets  up  internal  stresses  that  apparently  cause  an  increase 
in  strength. 

The  effect  of  increasing  the  thickness  of  the  metal  is  to  make 
a  considerable  reduction  in  width  on  the  supporting  girder.  Try 
a  ^-in.  steel  strap. 

5350 
k  =  oooo  —  TTEs  =  9.2  ins.,  of  which  each  leg  will  be  one-half, 

Zooo  X  U.O 

or  4.6  ins.  By  increasing  the  thickness  £  in.  the  width  of  the  strap 
has  been  reduced  nearly  one-half.  The  wide  strap  will  weigh 
10.4  Ibs.  per  lineal  foot.  The  narrow  strap  of  thicker  metal  will 
weigh  7.8  Ibs.  per  lineal  foot,  so  will  be  the  cheaper  strap  to  use. 

If  a  stirrup  is  not  designed  to  be  safe  according  to  calculations 
such  as  those  illustrated  it  should  not  be  used.  A  lack  of  strength 
in  bending  is  sometimes  claimed  to  be  taken  care  of  by  using  a 
longer  support  and  holding  it  down  with  lag  screws  or  spikes. 


96 


PRACTICAL  STRUCTURAL  DESIGN 


The  longer  support  increases  the  bending  moment  and  the  holding 
down  strength  of  the  fastenings  must  be  investigated.  The  leg 
is  sometimes  run  across  the  top  of  the  girder  and  bent  down  on 
the  other  side  and  there  fastened,  which  is  sometimes  good,  but  the 
increase  in  material  added  to  the  cost  of  fastenings  and  the  cost 
of  labor  to  drive  them  amounts  to  more  than  the  cost  of  the  addi- 
tional thickness  necessary  to  prevent  straightening.  A  stirrup 
strong  enough  to  carry  a  load  without  bending  is  more  satisfactory 
than  one  confessedly  weak  with  which  fastenings  must  be  used. 

Tops  of  beams  and  girders  should  not  be  cut  to  make  a  seat 
for  stirrups.  This  weakens  the  timber,  so  the  under  side  of  the 
floor  planking  should  be  cut  to  make  pockets  for  the  stirrups.  A 
cheaper  method  is  to  lay  a  strip  of  wood  to  carry  the  flooring  on 
top  of  the  beam  between  stirrups.  When  the  floor  is  double  the 
under  layer  may  be  cut  away  at  the  stirrups,  the  upper  layer  being 
amply  strong  to  carry  over  the  small  hole.  Fig.  62  is  reproduced 
from  "Ryerson's  Ready  Reference."  The  stirrups  illustrated  are 
made  of  wrought  iron  and  the  recommendation  is  made  in  the  book 
that  the  following  sizes  mentioned  in  the  table  should  in  general 
be  used  for  the  size  of  joist  supported,  the  stirrups,  unless  other- 
wise specified,  being  furnished  \  in.  smaller  than  nominal  size  of 
timber  or  joist.  Wall  hangers  rest  on  plates  as  shown. 

TABLE  OF  STIRRUP  SIZES  AND  CAPACITIES 


Size  of  Joist  or  Timber 
to  be  Supported 

Section  of  Stirrup 

Capacity  of  Stirrup 

2  x    8  to    3  x  10 
4  x  10  to    4  x  12 
6  x  12  to    3  x  14 
8x12  to    4x14 
6x  14 
8  x  14  to  10  x  14 

1x3 

|X4 

7,500  Ibs. 
11,250  Ibs. 
13,500  Ibs. 
21,000  Ibs. 
24,000  Ibs. 
30,000  Ibs. 

Another  method  for  carrying  the  ends  of  joists  on  a  girder  when 
head  room  is  to  be  saved  and  the  joists  cannot  rest  on  top  of  girders 
is  shown  in  Fig.  63.  This  depends  upon  shearing  resistance  of 
the  spikes.  First  find  the  width  of  bearing  required  for  each  joist 
by  dividing  the  reaction  by  the  bearing  strength  across  the  grain. 
Use  nails  having  a  length  practically  three  times  the  thickness 
of  the  bearing  strip  as  a  minimum,  so  they  will  go  into  the  girder 


PROBLEMS  IN  DESIGN  OF  BEAMS 


97 


a  depth  about  twice  the  thickness  of  the  bearing  strip.  The  num- 
ber of  nails  to  use  depends  on  the  reaction,  and  the  thickness  of  the 
nail.  Divide  the  reaction  in  pounds  by  100  to  get  the  number  of 
20d.  nails ;  by  150  for  30d.  nails ;  by  175  for  40d.  nails ;  by  200 
for  50d.  nails;  by  225  for  60d.  nails.  There  is  considerable  dif- 
ference in  weight  between  nails  and  spikes  having  the  same  desig- 
nation and  the  above  figures  refer  to  nails.  The  nails  should  be 
spaced  at  least  3  ins.  apart  horizontally  and  this  can  be  accom- 
plished by  putting  half  near  the  bottom  of  the  strip  and  half  near 
the  top,  thus  staggering  them.  The  size  of  nail  to  use  will  therefore 
be  determined  by  the  spacing  when  the  reaction  is  considerable. 


Fig.  63  —  Wood  End  Bearings  for  Joists 

The  above  described  bearing  strip  support  for  joists  is  a  cheap 
method.  Formerly  it  was  customary  to  use  girders  considerably 
larger  than  were  necessary  and  seats  were  cut  into  them  for  the 
joists.  This  increased  the  labor  cost  and  when  water  settled  into 
the  joints  they  rotted.  The  introduction  of  slow  burning  construc- 
tion also  acted  to  throw  the  gaining  of  joists  into  girders  into 
disrepute  because  of  the  increased  fire  risk  in  the  joints.  When 
a  nailed  bearing  strip  is  used  it  should  be  carefully  computed. 
The  bearing  should  be  at  least  half  an  inch  wider  than  the  com- 
puted bearing.  The  ends  of  the  joists  should  be  carefully  fitted. 
It  is  advisable  with  thick  joists  to  top  nail  them  to  the  girder  to 
prevent  twisting  or  winding.  With  thin  joists  a  solid  bridging 
should  be  inserted  at  the  ends,  nailed  to  the  girder.  When  this  is 
done  many  of  the  objections  to  the  joiners  pocket  are  introduced. 
Therefore  metal  hangers  are  better  when  for  any  reason  it  is  not 
advisable  to  have  the  joists  rest  on  top  of  a  girder.  In  slow  burn- 
ing construction  neither  hangers  or  bearing  strips  are  proper. 
The  thickness  of  joists  in  such  construction  should  be  not  less 
than  half  the  depth  and  the  minimum  cross-sectional  area  should 
be  72  inches.  All  joists  should  rest  on  top  of  girders. 

The  maximum  shear  on  a  wooden  beam  is  along  the  neutral 
axis  and  season  checks  are  apt  to  occur  here,  so  nailing  strips 


98  PRACTICAL  STRUCTURAL  DESIGN 

should  be  as  far  from  the  neutral  axis  as  possible,  which  indicates 
the  bottom  of  the  girder,  as  a  proper  location.  When  a  seating  is 
cut  into  the  bottom  of  a  joist  it  is  apt  to  cause  the  joist  to  split, 
and  if  season  checks  open  at  this  point  the  joist  will  be  greatly 
weakened.  Half  the  depth  of  a  joist  may  be  cut  out  at  the  very 
end  without  weakening  it  for  carrying  purposes,  for  the  bending 
moment  at  the  end  is  zero,  but  if  a  joist  splits  for  any  considerable 
distance  from  the  end  it  is  greatly  weakened.  The  principal 
objection,  therefore,  to  the  nailed  bearing  strip  is  the  danger  of 
splitting  the  joists  at  the  upper  edge  of  the  seat.  If  the  joists  rest 
without  cutting  on  the  bearing  strip  and  the  strip  is  properly  de- 
signed, it  cannot  be  condemned.  There  will,  however,  in  this  case, 
be  a  projection  of  the  girder  below  the  ceiling  equal  to  the  width 
of  the  strip. 

6.  Design  a  laminated  floor  to  carry  a  total  load  of  48  Ibs.  per 
square  foot  on  a  span  of  16  ft.  Use  white  pine  with  a  fiber  stress 
of  800  Ibs.  per  square  foot.  Ignore  deflection. 

A  laminated  floor  is  a  solid  floor  consisting  of  2-in.  planks  spiked 
side  by  side.  The  width  to  use  in  designing  is  12  ins.,  the  load 
being  in  pounds  per  square  foot. 

M  =  1.5  x  48  X  162  =  18,432  in.  Ibs. 
/  =  800  . ' .  R  =  800  +  6  =  133. 


133  X  12 

Use  2  ins.  X  4  ins.,  which  will  give  an  actual  depth  of  3f  ins. 
If  the  deflection  is  not  to  exceed  -5%-$  of  the  span,  the  deflection 

.„  ,     16  x  12      n  _„.  . 
will  be  —  =  0.534  in. 

ooU 

Z/2  162 

Actual  deflection  =  7^-  =  7^ — 0  _oe  =  1.535  ins. 
46n      46  X  3.625 

The  formulas  previously  given  for  deflection,  page  80,  are 
based  on  the  fiber  stress  and  to  avoid  several  trial  computations 
to  ascertain  the  depth  with  the  reduced  stress  use  the  rule  that : 

The  deflection  in  beams  varies  as  the  cube  of  the  length  in  feet 
divided  by  the  breadth  in  inches  multiplied  by  the  cube  of  the  depth 
in  inches. 

Expressed  as  a  formula  it  appears: 

Depth  varies  as  rrr 
bh3 


PROBLEMS  IN  DESIGN  OF  BEAMS 

Let  L  =  span  in  feet, 

6  =  breadth  in  inches, 
h  =  depth  in  inches  (required  for  strength), 
x  =  depth  in  inches  (required  for  deflection), 
D  =  deflection  found  in  inches, 
d  =  allowable  deflection  in  inches. 


Then  ,lDh3    .  Jl.535  x  3.G253      ,  ,-  . 

*  =  Shr  =  S/    0.534     =5-15ms- 

Use  nominal  2-in.  x  6-in.  planks. 
The  formula  is  developed  as  follows  : 

dU  _  PL* 

bh?       btf  ' 

which  becomes  ^!  x  M. 

bh3  x  DL3 

Cancelling  common  factors,  we  get  j^  and  y?  =  —  T-,  the  formula 

used  above. 

7.  Assuming  a  floor  with  same  load  and  fiber  stress  is  to  be 
carried  on  joists  find  the  size  required  for  joists  12  ins.  center  to 
center  and  16  ins.  center  to  center.  Deflection  to  govern. 

The  allowable  deflection  =  163^Q12  =  0.534  in. 
L2  162 


46Z)  46  X  0.534 
The  depth  in  this  example  needed  to  avoid  undue  deflection  is 
based  on  the  fiber  stress  used  in  design,  for  the  breadth  is  governed 
by  the  depth.  In  the  case  of  the  laminated  floor  a  constant  breadth 
of  12  ins.  was  used  and  the  deflection  was  fixed  by  a  lower  fiber 
stress  than  that  used  for  strength  only. 

The  bending  moment  for  a  width  of  1  ft.  =  18,432  in.  Ibs.  (from 
the  last  example).    The  thickness  of  the  joist  will  be 

18,432 

6  =  133  x  10.625*  =  L23mS- 

Use  nominal  1.5-in  x  11-in.  joists,  12  ins.  center  to  center. 
With  joists  spaced  16  ins.  center  to  center  the  bending  moment 
is  increased  one-third,  18,432  x  1.333  =  24,600  in.  Ibs. 

24,600 

6  =  133  X  10.625*  =  IM  mS' 
Use  nominal  2-in.  x  11-in.  joists,  16  ins.  center  to  center. 


100  PRACTICAL  STRUCTURAL  DESIGN 

The  increase  in  amount  of  lumber  is  about  one-fourth,  while 
the  increased  carrying  capacity  is  one-third,  so  it  will  be  better 
to  use  the  16-in.  spacing  than  to  use  the  12-in.  spacing.  Thinner 
joists  than  2-in.  are  not  advisable  when  it  is  possible  to  avoid  them, 
so  this  is  another  reason  for  using  the  wider  spacing.  To  make  the 
floors  stiff  and  to  avoid  bending  under  load,  or  warping  of  the 
joists  from  any  cause,  lines  of  cross  bridging  should  be  used  at 
intervals  approximately  twenty-four  times  the  thickness  of  the 
joist.  This  for  2-in.  joists  will  be  48-in.  (4-ft.)  centers. 

Beams  on  a  Slope 

Let     S  =  length  of  beam  on  slope, 
L  =  horizontal  span, 

W  =  total  load  uniformly  distributed  on  the  slope, 
W  =  total  load  beam  can  carry, 

TI7/  WS 

then     W  =  -jr- 
Li 

The  foregoing  applies  in  the  case  of  stringers  supporting  stairs 
and  inclined  rafters  carrying  a  load  on  the  upper  surface.  There 
is  a  horizontal  and  a  vertical  force  acting  when  a  beam  is  inclined 
and  the  resultant  thrust  increases  the  compression  and  decreases 
the  tension  in  the  fibers.  It  is  usually  unimportant  and  may  be 
neglected  when  the  slope  is  less  than  thirty  degrees,  but  should 
be  investigated  in  any  .case.  When  an  inclined  member  of  a  truss 
carries  a  load  on  the  upper  surface  in  addition  to  the  direct  thrust, 
the  member  must  be  designed  to  take  these  loadings  into  account. 
Let  M  =  bending  moment  due  to  the  load. 
/  =  fiber  stress. 

A  =  area  of  member  in  cross  section 
W  =  direct  load  along  axis  of  the  member. 
h  =  depth  of  the  member. 

I  =  moment  of  inertia  of  member,  which  is  assumed  here 
to  be  symmetrical. 

,          W      Mh 
then      /-  +  _±_ . 

In  the  above  expression  the  first  part  gives  the  average  fiber 
stress  due  to  the  direct  load  acting  along  the  length  of  the  member, 
that  is,  the  push.  The  second  half  is  the  familiar  expression  for 
the  fiber  stress  in  a  beam  bending  under  load.  Use  it  twice,  once 
with  a  positive  sign  and  once  with  a  negative  sign.  The  ex- 


PROBLEMS  IN  DESIGN  OF  BEAM  101 

pression  sometimes  appears, 

W     M 
f=+A±^' 
in  which  S  =  section  modulus. 

Buckling  of  Beams 

The  author  has  been  careful  in  calling  attention  to  the  fact  that 
beam  formulas  and  tables  of  carrying  capacity  of  beams  assume 
the  beams  to  be  stayed  for  lateral  stiffness.  If  a  beam  is  too  long 
the  upper  half  acts  as  a  slender  column  having  a  least  dimension 
equal  to  the  breadth.  When  a  floor  is  fastened  to  the  upper  sur- 
face along  the  length  it  is  usually  a  sufficient  stay.  It  is  best, 
however,  to  have  a  stay  as  well  for  the  lower  edge  of  the  beam. 
A  familiar  illustration  is  the  cross  bridging  between  wood  floor 
joists  placed  at  intervals  of  about  24  times  the  breadth  of  the  joist. 
In  steel  beams  the  lateral  stays  should  be  spaced  at  intervals  not 
exceeding  40  times  the  width  of  the  flange.  All  stays  prevent 
a  sidewise  buckling,  and  the  stay  at  the  lower  edge  prevents  a 
blow  from  pushing  the  beam  to  one  side,  which  would  cause  the 
loading  to  become  eccentric  and  thereby  increase  the  stresses. 
The  effect  of  lateral  deflection  and  eccentric  loading  is  to  set  up 
the  simultaneous  action  of  a  direct  thrust  plus  bending. 

Stiffness  of  Wood  Beams 

The  following  formula  was  evolved  by  Thomas  Tredgold,  a  noted 
British  authority  on  carpentry  in  the  last  century.  A  beam 
designed  according  to  this  formula  will  deflect  less  than  ^^  the 
span. 

6  =  breadth  of  beam  in  inches, 

h  =  depth  of  beam  in  inches, 

L  =  span  in  feet, 

P  =  concentrated  load  in  middle  of  span, 

W  =  uniformly  distributed  load  =  0.625P, 


3\UPC       AUWC 
k  =  \-F"  =  \-T~' 

UPC      UWC 
o  =  — r 


h* 

C  =  a  constant  =  0.010  for  fir  and  yellow  pine. 
=  0.013  for  oak  and  white  pine. 


CHAPTER  IV 
Girders  and  Trusses 

A  METHOD  frequently  used  by  carpenters  to  strengthen 
joists  and  beams  is  shown  in  Fig.  64.     Two  pieces  are 
nailed  as  indicated,  the  presumption  being  that  they  exert 
an  arching  action  because  the  ends  abut  at  the  middle  of  the  span 
and  the  nails  hold  the  pieces  in  place  when  thrust  is  exerted. 

™.        Wood    shrinks 
when  it  dries,  so 


Fig.  64  —  A  Poor  Method  for  Reinforcing  Joists 


the  close  contact 
is  lost,  and  then 
considerable  deflection  must  take  place  before  the  ends  again  meet, 
the  bending  being  sufficient  to  cause  failure  in  many  instances. 
Provided  the  hoped-for  arch  action  does  occur  there  will  be  such 
a  pushing  against  the  nails  that  the  wood  is  bound  to  split.  How- 
ever, assuming  the  arch  action  does  take  place  and  the  nails  do 
not  split  the  pieces  the  reinforcement  is  not  effective.  For  effec- 
tive arch  action  there  must  be  substantial  abutments  provided. 
If  there  are  no  substantial  abutments  a  tie  rod  is  necessary  to  tie 
the  ends  together  and  take  the  thrust.  There  being  no  tie  rod, 
it  is  evident  that  the  lower  part  of  the  joist  will  have  to  act  as 
a  tie.  We  know  that  when  a  beam  is  loaded  the  lower  fibers  are 
stressed  in  tension  and  the  upper  fibers  are  stressed  in  compres- 
sion. To  increase  the  tension  in  the  bottom  by  adding  to  it  the 
amount  required  to  take  care  of  the  thrust 
in  the  diagonal  reinforcing  strips  is  not 
helpful.  This  old-time  method  is,  therefore, 
based  on  a  fallacy  and  should  be  abandoned. 
In  Fig.  65  is  shown  another  method,  the 
reinforcing  being  spiked  along  the  top  edge 
to  make  a  beam  of  T-section.  This  raises  the 


Fig.  65  —  T-beam  of 
Wood 


neutral  surface  so  the  increased  area  in  compression  is  supposed 
to  be  offset  by  an  increased  area  in  tension. 

Assume  a  joist  3  ins.  X  14  ins.  of  wood  in  which  a  fiber  stress  of 
102 


GIRDERS  AND  TRUSSES  103 

1200  Ibs.  per  square  inch  is  used.    A  strip  1  in.  x  4  ins.  is  spiked  on 
each  side  along  the  top.    How  much  is  the  strength  increased? 

™        •  •     i    4      ^.u    *f       fib?      1200  x  3  X  142 

The  original  strength,  Mr  =  J-^-  =  -     —  -  --  =  117,600  in. 

Ibs.,  the  neutral  plane  being  in  the  middle  of  the  joist.   To  find  the 
position  of  the  neutral  plane  in  the  T-section  use  the  method  of 
moments,  taking  moments  about  the  lower  edge: 
The  original  piece,    3  X  14  x    7  =  294 
One  added  piece       1  x    4  x  12  =    48 
Second  added  piece  1  X    4  x  12  =    48 

390 

Area  =  (3  x  14)  +  (2x4)  =~~50  =  7'°8  ms< 
First  the  area  of  the  .beam  was  multiplied  by  the  distance  of 
the  center  of  gravity  above  the  bottom,  the  result  being  294. 
Then  the  area  of  each  added  piece  was  multiplied  by  the  distance 
of  its  center  of  gravity  above  the  bottom  of  the  beam.  This  was 
12  ins.,  being  half  the  depth  of  the  piece  added  to  the  difference  in 
depth  of  the  beam  and  the  piece.  The  products  were  added  to- 
gether, the  sum  being  390.  Dividing  by  the  total  area,  50  sq.  ins., 
the  distance  from  the  bottom  to  the  center  of  gravity  (center  of 
area  in  this  case)  was  found  to  be  7.08  ins. 

The  original  moment  arm  =  f  x  14  =  9.333  ins.,  that  is,  9.333  ins. 
Before  the  pieces  were  added  at  the  top  the  moment  of  resistance 
was  equal  to  the  area  on  one  side  of  the  neutral  axis  multiplied  by 
the  average  fiber  stress  times  the  moment  arm,  that  is: 


~-  x  3  x  7  x  9.333  =  117,600  in.  Ibs. 

The  strength  of  the  beam  is  fixed  by  the  fiber  stress  and  the 
smaller  stressed  area.  In  this  T-section  the  smaller  area  is  the 
portion  in  tension  below  the  neutral  axis  and  the  resisting  moment 

=  —  X  3  x  7.08  X  9.333  =  118,944  in.  Ibs. 

a 

The  increase  in  strength  is  very  small,  so  the  area  added  above 
the  neutral  axis  was  excessive.  Better  results  would  have  been 
obtained  by  nailing  one  strip  along  the  bottom  and  one  along 
the  top,  thus  increasing  the  area  equally  in  tension  and  compres- 
sion, without  altering  the  position  of  the  neutral  axis.  The  proper 
method  to  follow  is  to  increase  the  thickness  by  adding  boards 
on  one  or  both  sides  for  the  full  depth.  An  example  will  be  worked 
out: 


104  PRACTICAL  STRUCTURAL  DESIGN 

In  a  mill-constructed  building  7  ins.  x  14  ins.  white  pine  beams 
spaced  5  ft.  on  centers  are  used  with  a  span  of  20  ft.  The  allow- 
able maximum  fiber  stress  is  800  Ibs.  per  square  inch  and  the  beams 
are  to  be  strengthened  so  the  total  floor  load  can  be  increased  to 
100  Ibs.  per  square  foot,  inclusive  of  floor,  beams,  and  live  load. 

Testing  first  the  strength  of  the  beams  against  failure  by  longi- 
tudinal shear  on  the  neutral  axis,  the  unit  shear  being  one-tenth 
the  allowable  fiber  stress, 

4X7X14  x  80 


The  total  panel  load  will  be  5  x  20  X  100  =  10,000  Ibs.,  so  the 
beam  will  carry  the  additional  load  without  failing  in  shear. 

Mb  =  1.5  x  10,000  X  20  -  300,000  in.  Ibs. 
M,  =  80°XJX142  =  182,933  in.  Ibs. 


Then  the  difference  between  the  bending  moment  and  the 
resisting  moment  is  300,000  -  182,933  =  117,067  in.  Ibs.,  which 
difference  must  be  cared  for  by  reinforcement.  To  secure  equal 
deflection  the  reinforcement  should  be  the  same  wood,  white  pine, 
but  the  difference  will  not  be  appreciable  in  this  case,  and  to  use 
yellow  pine  will  give  a  smaller  piece  for  reinforcement  because 
of  the  higher  allowable  fiber  stress.  The  beam  is  in  an  old  build- 
ing and  quite  likely  the  maximum  deflection  in  the  white  pine  has 
been  reached,  and  there  is  a  decided  permanent  set.  The  reinforce- 
ment should  be  added  when  the  floor  is  unloaded  in  order  to  enable 
the  old  beam  and  the  new  pieces  to  deflect  together  when  the 
live  load  is  added,  the  difference  in  deflection  between  the  two  kinds 
of  wood  being  cared  for  by  the  deflection  due  to  dead  load  in  the 
wood  having  the  greatest  deflection. 

Assuming,  therefore,  yellow  pine  with  a  fiber  stress  of  1300  Ibs. 
per  square  inch  and  a  depth  of  .14  ins.  the  thickness  is  to  be  com- 
puted. Let 

R  =  1300  •*-  6  =  217 
,  M  117,067 
6  =  ^  =  217tl4l=2J6mS- 

Use  two  If-in.  planks,  one  on  each  side.  When  surfaced  the 
thickness  will  be  practically  2f  ins. 

The  load  is  uniformly  distributed;  the  original  beam  is  large 
enough  to  carry  the  required  load  without  a  shearing  failure; 


GIRDERS  AND  TRUSSES  105 

the  diagram  for  bending  moment  due  to  a  uniformly  distributed 
load  is  a  parabola;  therefore  the  reinforcing  planks  need  not  ex- 
tend the  full  length  of  the  beam.  They  would  have  to  extend  the 
full  length  if  there  were  danger  of  a  longitudinal  shearing  failure, 
and  the  thickness  of  the  reinforcement  would  also  be  governed 
by  the  requirement  for  shear. 

Dividing;  182,933  -=-  300,000  =  0.61,  which  shows  that  the  resist- 
ing moment  of  the  beam  is  61  per  cent  of  the  bending  moment 
created  by  the  load.  The  ends  of  the  reinforcing  planks  must  extend 
each  side  of  the  middle  of  the  span  to  the  point  where  the  bend- 
ing moment  is  61  per  cent  of  the  bending  moment  at  the  middle 
of  the  span.  This  may  be  obtained  graphically  by  constructing 
a  parabola  with  a  base  equal  to  20  and  a  height  about  equal  to 
this,  the  height  divided  decimally  to  any  scale.  At  a  height  equal 
to  61  on  the  middle  ordinate  draw  a  horizontal  line  to  intersect 
the  parabola.  From  the  point  of  intersection  drop  a  perpendicular 
to  the  base.  This  defines  the  point  where,  theoretically,  the  rein- 
forcement may  end.  Practically  it  should  extend  a  little  further. 

The  lengths  of  the  reinforcing  planks  may  be  calculated  by  men 
who  can  solve  a  quadratic  equation.  The  bending  moment  on  a 
uniformly  loaded  beam  at  any  point  distant  x  from  one  end  is  as 
follows:  wLx  WX2 

ar,-_  — . 

Substitute  the  values  for  Mx,  w,  and  L  and  solve  for  x. 
W  =      '        =  523  Ibs.  per  lineal  foot  (in  even  numbers). 

-I  QO  OQQ 

MX  =  — -^ — •  =  15,244  ft.  Ibs.  (in  even  numbers).    This  moment 
is  the  resisting  moment  of  the  beam  without  reinforcement. 
Then  15,244  -  523  *  2°*  -  ™*. 

Clearing  of  fractions, 

2  x  15,244  =  30,488  =  523  x  20x  -  523z2. 
Dividing  by  the  coefficient  of  z2, 

58.3  =  20x  -  x2. 

Transposing,      x2  -  20x  =  -58.3. 
Extracting  the  square  root, 


+  -TT  ±  V    ir    -  58.3  =  3.55  ft. 


106  PRACTICAL  STRUCTURAL  DESIGN 

The  reinforcing  planks  (theoretically)  should  have  a  length 
of  20  -  (2  x  3.55)  =  13.90  ft.  Practically  it  will  be  best  to  make 
them  15  ft.  long,  which  leaves  2  ft.  6  ins.  without  reinforcement  at 
each  end  of  the  beam. 

The  planks  must  be  attached  to  the  beam  by  screws  or  nails, 
the  latter  being  the  cheaper.  To  get  the  best  results  the  length 
must  be  not  less  than  three  times  the  thickness  of  the  plank,  in 
order  that  the  nail  may  be  embedded  in  the  beam  a  depth  at  least 
twice  the  thickness  of  the  plank.  From  a  table  of  sizes  of  standard 
steel  wire  nails  and  spikes  (in  the  steel  manufacturers'  handbooks) 
we  find  a  30d.  nail  is  4.5  ins.  long,  the  length  required.  There  must 
be  enough  nails  used  so  the  beam  and  planks  will  act  together 
and  the  force  to  be  resisted  is  shear,  for  if  the  beam  bends  and  the 
planks  do  not  bend  there  will  be  a  sliding  movement  between 
them. 

When  a  nail  resists  a  shearing  force  three  actions  are  set  up: 
1.  A  bending  caused  by  the  pull  of  one  piece  against  the  nail  em- 
bedded in  the  other  piece.  2.  A  shear  hi  the  nail  which  is  caused 
if  the  nail  is  so  stiff  that  it  will  not  bend.  3.  Bearing  against  the 
wood  in  which  the  nail  is  embedded.  The  size  of  the  nail  must  be 
proportioned  to  care  for  the  action  most  likely  to  cause  a  failure. 
When  the  material  to  be  held  is  wood  the  bearing  action  of  the  nail 
against  the  wood  is  the  only  one  to  be  considered,  for  if  the  nail 
furnishes  area  enough  to  transmit  the  shear  it  will  be  thick  enough 
to  resist  bending  and  also  thick  enough  not  to  shear  across.  Rivets 
in  metal  have  to  be  similarly  proportioned,  but  bending  is  seldom 
feared  while  failure  by  shear  of  the  rivet  or  by  insufficient  bear- 
ing against  the  metal  is  practically,  and  usually,  of  equal  impor- 
tance. Both  must  be  figured,  whereas  in  the  case  of  wood  only 
the  bearing  value  is  considered.  The  bearing  value  is  computed 
as  follows: 

The  30d.  nail  (not  spike)  is  made  from  No.  5  wire,  the  diameter 
being  0.207  in.  The  cross-sectional  area  through  a  l|-in.  plank 
is  0.207  x  1.5  =  0.311  sq.  ins.  The  compressive  value  of  the  softer 
wood  must  be  used,  which  is  700  Ibs.  per  square  inch  with  the  grain, 
assuming  the  nail  to  bear  on  the  end  of  the  wood  where  it  enters. 
The  bearing  value  for  one  nail  is  found  by  multiplying  the  bearing 
area  by  the  allowable  fiber  stress  in  compression  with  the  grain, 
0.311  x  700  =  218  Ibs.  This  is  the  method  to  be  used  when  no 
data  is  at  hand  giving  the  actual  safe  bearing  values. 


GIRDERS  AND  TRUSSES 


107 


The  actual  safe  bearing  value  for  any  nail  is  about  two-thirds 
of  the  value  as  above  computed.  One  reason  is  that  there  is  no 
common  gauge  used  by  nail  makers,  so  that  while  tables  may  show 
that  nails  are  made  of  certain  wire,  not  all  tables  give  the  diameter 
of  the  wire  in  decimals  of  an  inch,  and 
there  being  a  number  of  wire  and 
metal  gauges  in  use  we  do  not  know 
the  exact  sizes  of  the  nails  used  in  the 
published  experiments.  The  experi- 
ments referred  to  may  have  been 
made  with  nails  not  quite  so  thick  as 
the  nails  used  in  computing  the  bear- 
ing value.  A  second  reason  for  the 


Fig.  66 — Shear  Diagram  for 
Original  Beam 


actual  bearing  value  being  so  small  is  that  the  nails  push  the  fibers 
of  the  wood  aside  and  start  a  splitting  action,  which  is  increased 
when  the  shearing  action  is  set  up.  This  second  reason  is  no  doubt 
much  more  important  than  the  first.  The  method  of  figuring 
bearing  value  just  illustrated  is  correct  for  bolts  for  which  holes 
must  be  bored,  but  gives  a  value  about  50  per  cent  too  large  for 
driven  wire  nails  and  for  screws.  The  designer  must  not  forget 
this.  Having  settled  on  the  size  of  nail  and  the  bearing  value  of 
each  nail,  the  number  and  spacing  must  be  determined. 

Fig.  66  is  the  shear  diagram  for  the  uniformly  loaded  beam. 
At  each  end  the  shear  =  reaction  =  10,450  -=-  2  =  5225  Ibs.    The 

nails  should  be 
closer  together 
near  the  ends, 
where  the  shear 


Fig.  67  —  Shear  Diagram  for  Reinforced  Beam 


is  a  maximum, 
so  theoretically 
the  spacing 
should  vary 

from  nail  to  nail.  Practically  the  spacing  can  be  maintained  at 
uniform  intervals  for  each  foot,  which  makes  the  diagram  resemble 
Fig.  67,  the  reinforcement  ending  2.5  ft.  from  each  end. 

The  method  to  be  described  follows  the  common  method  for 
spacing  rivets  in  the  flanges  of  plate  girders.  There  is  another 
method  which  will  later  be  illustrated,  because  it  shows  exactly  how 
the  stresses  in  the  top  and  bottom  flanges  of  plate  girders  affect  the 
rivets  used  to  connect  the  flanges  to  the  web.  In  the  present 


108  PRACTICAL  STRUCTURAL  DESIGN 

example  the  original  beam  may  be  said  to  represent  the  web, 
the  reinforcing  pieces  the  flange  members,  and  the  nails  the  rivets 
used  in  plate  girders.  The  spacing  of  stirrups  in  a  reinforced-con- 
crete  beam  is  another  illustration  of  the  transmission  of  stresses 
from  one  part  of  a  beam  to  another  by  designing  to  resist  shear. 
In  a  truss  the  sizes  of  the  members  are  varied,  for  the  panel  lengths 
are  equal.  When  the  nails  in  a  reinforced  wooden  beam  are  of  the 
same  size,  the  rivets  in  a  plate  girder  are  the  same  size  and  the 
stirrups  in  a  reinforced-concrete  beam  are  the  same  size,  the 
variation  in  shear  is  taken  care  of  by  varying  the  spacing. 

COOK  v  8 

Two  feet  from  the  end  of  the  beam  the  shear  is  —  —  —  =  4180 

Ibs.    The  width  of  the  original  beam  is  7  ins.,  and  the  two  planks 

4180  x  7 
increase  the  width  to  9.75  ins.,  or  —  ^-^=  —  =  3000  Ibs.,  which  will 


.       .      4180  -  3000 
be  carried  by  the  original  beam,  leaving  -  •=  --  =  590  Ibs.  to 

be  carried  by  each  plank.    The  nails  must  transfer  this  from  the 
beam  to  the  plank  and  they  should  be  driven  1  in.  from  the  edge, 
both  top  and  bottom.     Let 
V  =  total  vertical  shear  at  the  point  considered, 
r  =  resistance  of  one  nail  (bearing  value), 
d  =  distance  hi  inches  between  lines  of  nails   (in  the  present 

example  d  =  12  ins.  vertically), 

p  =  pitch  of  nails  in  inches  (the  horizontal  distance  center  to 
center  between  nails)  ; 

rd      145  X  12      0  n_  . 
then  p  =  —  =  —  —  —  =  2.95  ms. 

Space  the  nails  2f  ins.  center  to  center  along  the  upper  and  lower 
edge  for  at  least  6  ins.  at  each  end,  the  first  nail  being  driven  1  in. 
from  the  end  of  the  plank. 

COOK,    v   7 

Shear  3  ft.  from  erid  =        ^      :=  3660  Ibs. 

.     ,       3660  -  3000 
Shear  carried  by  each  plank  =  --  •=  --  =  330  Ibs. 

p  =  —  —  —  —  =  3.92  ins.    Space  the  nails  3f  ins.  center  to  center 
along  the  upper  and  lower  edge  of  the  plank  for  1  ft. 
Shear  4  ft.  from  end  =  522^QX  6  =  3135  Ibs. 


GIRDERS  AND  TRUSSES  109 

Shear  carried  by  each  plank  =  3135  ~  300°  =  67.5  ibs. 

145  x  12      0_  _  . 

P=-67^-=25-8m- 

Nails  should  be  driven  not  more  than  12  ins.,  on  centers,  so, 
beginning  at  the  end  of  the  fourth  foot  from  the  end  of  the  beam 
drive  nails  on  12-in.  centers  top  and  bottom.  Along  the  neutral 
axis  drive  nails  on  18-in.  centers.  The  completed  work  is  shown 
in  Fig.  68. 

No  reduction  in  area  was  figured,  as  nails  merely  push  wood 
fibers  aside,  but  when  bolts  are  used  the  effective  depth  of  the 
beam  is  reduced  by  the  thickness  of  each  line  of  bolts.  If  bolts 


Bottom 


Fig.  68  —  Beam  Reinforced  by  Planks  on  the  Sides 

f  in.  in  diameter  are  used  in  two  lines,  and  the  hole  for  each  bolt 
is  |  in.,  the  effective  depth  is  reduced  by  2  x  |  =  If  ins.  This  is 
serious,  for  the  strength  of  beams  varies  with  the  squares  of  the 
respective  depths. 

Sometimes  beams  are  reinforced  by  nailing  a  plank  or  strip  of 
steel  along  the  bottom.  Assume  the  same  conditions  as  in  the 
last  example,  and  use  a  thin  white  pine  plank  on  the  bottom. 
Maintaining  the  breadth  the  problem  is  to  obtain  a  new  depth. 

The  fiber  stress  for  white  pine  is  800  Ibs.,  so  R  =  800/6  =  167 


and  h  =  y         =  Y>y  =  16  ins.    The  original  depth  is  14 

ins.,  so  a  plank  2  ins.  thick  by  7  ins.  wide  must  be  spiked  or 
bolted  to  the  bottom.  A  thick  plank  like  this  must  be  fastened 
with  bolts,  and  the  holes  will  reduce  the  area,  which  will  make 
necessary  an  increase  in  thickness.  Methods  for  finding  the  length 
of  the  reinforcing  plank  and  the  pitch  of  the  bolts  have  been  given, 
the  depth  used  being  the  full  depth  of  the  original  beam  plus  half 
the  thickness  of  the  reinforcing  plank. 


110 


PRACTICAL  STRUCTURAL  DESIGN 


To  reinforce  with  a  steel  plate  on  the  bottom  use  a  moment 
arm  =  f  X  14  =  9.333  ins.  The  fiber  stress  in  the  steel  will  be 

16,000  Ibs.  per  square  Inch.     Then  area  of  plate  =     ool^'^Jnnn 

9.000  x  ib,ooo 

=  0.783  sq.  ins.,  and  0.783  -=-  7  =  0.118  ins.,  the  thickness  of  the 
plate.  Use  lag  screws  to  fasten  the  plate  to  the  beam,  the  proper 
pitch  being  determined  as  in  the  last  example,  using  the  full  depth. 
The  objection  to  the  use  of  the  steel  plate  is  that  the  compression 
in  the  upper  half  of  the  beam  is  increased,  although  the  effect 
of  adding  the  plate  is  to  lower  the  neutral  plane.  The  proper 
method  for  reinforcing  a  beam,  or  girder,  in  place  is  to  add  planks 
on  one  side  or  on  both  sides,  but 
when  fixtures  or  wires  are  in  the 
way  it  may  be  best  to  use  a  steel 
plate  on  the  bottom. 

Compound  beams  have  been  made 
consisting  of  two  shallow  beams 
superimposed  (Fig.  69).  If  not  care- 
Fig.  69  —  Compound  Beam  fully  fastened  together  they  act 
singly,  because  the  line  between  them  is  in  the  position  occupied 
by  the  neutral  axis  of  a  solid  beam,  having  a  depth  equal  to  the 
combined  depth  of  the  two  pieces.  Several  methods  have  been 
used  to  cause  the  two  pieces  to  act  together,  one  of  which  is 
shown -in  Fig.  69,  the  other  in  Fig.  70. 

No  matter  how  thoroughly  the  pieces  are  fastened  together 
the  strength  of  such  a  compound  beam  is  only  about  70  to  75 
per  cent  of  the  strength  of  a  beam  of  equal  dimensions  made 
from  one  piece  of  timber.  The  deflection  of  such  a  beam  under 
load  is  much  greater  than  the  deflection  of  a  beam  of  equal  dimen- 
sions made  from  one  piece  of  timber. 

The  diagonal  side  pieces  shown  in  Fig.  69  should  be  preferably  of 
a  harder  wood  than  the  beam,  and  each  should  be  not  less  than  one- 
eighth  the  thickness  of  the  beam,  thus  making  a  beam  25  per  cent 
wider  than  the  width  of  the  pieces  of  which  it  is  composed.  The 
pieces  should  be  diagonal  and  slope  in  opposite  directions  on  the 
sides  of  the  beam.  Plenty  of  nails  must  be  used. 

In  Fig.  70  the  pins  should  be  of  hard  wood  or  of  metal.  It  is 
best  to  use  two  pieces  in  each  hole,  wedge-shaped,  so  they  may  be 
driven  tight  and  have  a  bearing  against  the  wood  the  full  width 
of  the  beam.  The  shear  being  greatest  along  the  neutral  axis, 


GIRDERS  AND  TRUSSES  111 

it  is  here  the  pieces  should  join  and  the  pins  be  driven.    Between 

the  pins  should  be  vertical  bolts  with  larger  washers  to  hold  the 

pieces  together.    The  spacing  of  the  pins  will  be  determined  in 

like  manner  as  the  pitch  of  nails  is  determined  when  reinforcing 

planks  are  used  on  the  side.    First  determine  the  bearing  value 

of  the   wood   and   the 

shearing  value  with  the 

grain.   Divide  the  shear 

where  a  pin  is  placed  by 

the    allowable    bearing  Fig.  70  -  Compound  Beam 

times  the  breadth  to. obtain  the  depth  of  the  hole,  half  of  which 
will  be  cut  in  each  half  of  the  beam.  The  shear  divided  by  the 
breadth  times  the  allowable  unit  shear  with  the  grain  gives  the 
minimum  distance  allowable  between. pins.  When  the  computa- 
tions are  completed  it  will  be  discovered  that  the  pins  get  farther 
apart  as  the  middle  of  the  span  is  approached. 

Flitch  plate  girders,  Fig.  71,  are  seldom  used  to-day,  although 
very  popular  at  one  time.  The  only  reason  for  referring  to  this 
type  of  compound  girder  here  is  to  show  wherein  it  fails.  A  flitched 
girder  consists  of  a  plate  of  steel,  or  wrought  iron,  between  two 
planks,  the  whole  construction  being  firmly  bolted  together. 
The  writer,  in  wrecking  old  buildings,  found  a  number  of  such 
beams  evidently  put  together  on  a  basis  of  relative  fiber  stresses, 
with  no  thought  for  relative  deflections.  He  worked  once  in  the 
office  of  an  architect  who  tried  to  get  him  to  design  such  a  beam 
in  this  way  and  the  man  was  greatly  surprised  when  the  proper 
method  was  shown  to  him.  The  method  used  is  as 
follows:  Assuming  a  maximum  bending  fiber  stress  of 
1300  Ibs.  per  square  inch  for  wood  and  16,000  Ibs.  per 
square  inch  for  steel,  the  relative  areas  of  wood  and 
steel  will  be  16,000  +  1300  =  12.5,  or  a  |-in  steel  plate 
between  two  |-in.  planks  makes  a  girder  having  the 
strength  of  four  f-in.  planks. 
Referring  to  the  deflection  formulas  it  is  seen  that  for  a  fiber 
stress  of  16,000  Ibs.  in  steel  the  deflection  in  inches  on  any  span 

=  ^rr,  while  for  a  fiber  stress  of  1300  Ibs.  in  wood  the  de- 

v)U/i 

flection  =  — —  Therefore  yellow  pine  deflects  60  -^  41  =  1.46 
times  as  much  as  steel,  under  the  respective  fiber  stresses  given. 


112  PRACTICAL  STRUCTURAL  DESIGN 

This  question  of  deflection  does  not  take  into  consideration  the 
thickness  of  the  material,  for  deflection  is  governed  by  the  span 
and  the  depth. 

The  statement  about  relative  deflections  means  that  if  the 
thicknesses  are  proportioned  by  the  relative  stresses,  then  the  wood 
planks  must  be  1.46  times  as  deep  as  the  steel  plate  between  them. 
This  will  not  do  in  practice,  so  it  is  necessary  to  obtain  the  rela- 
tion between  the  stresses  when  the  plate  has  a  depth  equal  to  the 
depth  of  the  inclosing  planks.  The  fiber  stress  in  the  wood  divided 
by  the  fiber  stress  in  the  steel  must  equal  the  modulus  of  elasticity 
of  the  wood  divided  by  the  modulus  of  elasticity  of  the  steel; 
that  is, 

fj!>_E* 

fs  ~  E.' 

/«EW      16,000  x  1,500,000 
Then'»=V=          30,000,000         =  800  Ibs.  per  square  inch. 

This  is  a  low  stress  for  yellow  pine,  so  a  softer  wood  can  be 
used.    Assume  a  wood  having  a  modulus  of  elasticity  of  1,000,000, 
iU       ,        16,000  x  1,000,000      coc  ., 
then/"=         30,000,000         =  535  Ibs.  per  square  inch. 

If  a  steel  fiber  stress  of  18,000  Ibs.  per  square  inch  is  assumed, 
the  fiber  stress  in  the  wood  =  600  Ibs.  per  square  inch.  The  com- 
putations show  that  for  a  flitch  plate  beam  a  soft,  cheap  wood  is 
the  kind  to  use.  It  is  wasteful  to  use  a  wood  in  which  a  high 
fiber  stress  may  be  permitted. 

To  design  a  flitched  girder  the  fiber  stresses  are  first  found. 
Then  assume  the  depth  and  thickness  of  the  steel  plate.  Find  how 
much  it  will  carry  as  a  thin  deep  beam  and  deduct  this  load  from 
the  total  load  to  be  carried.  The  difference  is  to  be  carried  by  the 
two  wood  planks  of  which  we  know  the  depth  and  the  fiber  stress, 
so  it  is  easy  to  find  the  thickness.  The  bolts  are  figured  to  trans- 
mit the  shear.  It  is  an  interesting  exercise  to  design  a  flitch  girder, 
but  a  rolled  steel  beam  or  a  trussed  wooden  girder  will  usually  be 
cheaper. 

Plate  Girders 

Plate  girders  are  compound  girders  made  of  wrought  iron  or 
steel,  the  latter  material  being  generally  used  to-day,  for  it  may 
be  used  with  a  higher  fiber  stress,  thereby  reducing  the  weight. 
When  rolled  beams  are  not  obtainable  in  a  large  enough  size  a 


GIRDERS  AND  TRUSSES 


113 


plate  girder  is  used,  provided  a  rolled  beam  cannot  be  made  to 
serve,  by  attaching  plates  to  the  flanges.  Tables  of  plate  girders 
are  given  in  the  steel  handbooks,  so  the  architect  or  builder  finds 
it  as  easy  to  select  a  plate  girder  for  much  of  his  work  as  it  is  to 
select  a  rolled  I-beam  or  channel  for  light  loads  on  shorter  spans. 
When  the  load,  or  the  span,  either  or  both,  make  a  plate  girder 
too  heavy,  a  trussed  girder  is  used. 

Fig.  72  shows  a  plate  girder.  The  thin  vertical  plate  is  known 
as  the  web  and  is  made  thick  enough  to  carry  the  shear.  It  acts 
also  as  a  long  slender  column,  so  must  be  safe  against  crippling. 
When  proportioned  to  carry  the  shear  and  the  thickness  is  greater 
than  sV  the  depth  between  the  rivets  in  the  upper  and  lower  flanges, 


Fig.  72  —  Plate  Girder  with  Cover  Plates  and  Stiffeners 

the  plate  is  safe  against  crippling.  When  designed  to  carry  the 
shear  and  a  thickness  less  than  ?\  the  depth  is  obtained,  it  is  neces- 
sary to  use  stiffeners  spaced  regularly  at  intervals  equal  to  the 
depth  of  the  girder.  Additional  stiffeners  are  placed  under  con- 
centrated loads  and  at  the  ends.  Intermediate  stiffeners  are  some- 
times crimped  over  the  flange  angles,  but  it  is  as  common  to  have 
fillers  placed  under  them,  the  thickness  of  the  fillers  being  equal 
to  the  thickness  of  the  flange  angle,  so  the  stiffeners  will  be  straight 
and  have  the  ends  resting  on  the  outstanding  leg  of  the  flange 
angles.  No  scientific  rules  seem  to  be  commonly  accepted  for 
designing  intermediate  stiffeners,  the  usual  empirical  method 
being  to  have  the  outstanding  leg  equal  in  width  to  -jV  the  depth 
plus  2  in. 

End  stiffeners  act  as  columns  to  carry  the  end  shear,  which  is 
delivered  to  them  by  the  web  plate  and  carried  to  the  bearing 
plate.  Stiffeners  under  concentrated  loads  are  designed  as  columns. 
The  ends  of  all  stiffeners  designed  as  columns  should  be  milled 
or  ground  to  fit  perfectly  against  the  bottom  flange  angles. 


114  PRACTICAL  STRUCTURAL  DESIGN 

Divide  the  load  by  a  fiber  stress  of  12,500  Ibs.  per  square  inch  for 
the  area  of  the  stiffeners  and  use  fillers  to  keep  the  stiffeners 
straight. 

The  flange  may  be  made  solely  of  angles  extending  the  whole 
length  of  the  web  plate,  or  of  angles  with  plates  riveted  to  them, 
the  latter  type  being  adopted  when  angles  alone  will  not  be  suffi- 
ciently strong.  The  plates  seldom  extend  the  full  length  and  if 
more  than  one  flange  plate  is  used  the  outer  plates  are  very  short, 
the  lengths  increasing  progressively  as  they  get  closer  to  the  angles. 
These  plates  are  known  as  cover  plates,  and  when  different  thick- 
nesses are  used  the  thinner  plates  are  on  the  outside. 

The  resisting  moment  is  determined  as  follows:  one-eighth 
the  area  of  the  web  is  considered  as  forming  part  of  the  flange. 
This  is  the  usual  custom,  but  some  engineers  use  only  TV  and 
some  T*J. 


*,    t      u 

Mr  of  web  = 

o 


b  =  thickness  of  web  plate. 
d  =  total  depth  of  plate. 

/  =  unit  fiber  stress  (usually  16,000  Ibs.  per  sq.  in). 
Mr  of  angles  =  Adf. 

A  =  area  in  sq.  ins.  of  the  two  angles  on  one  edge  of 

plate. 

d  =  distance  center  to  center  of  gravity  of  the 
angles  on  upper  and  lower  edges  of  web 
plate. 

/  =  unit  fiber  stress. 
Mr  of  cover  plates  =  Adf. 

A  =  area  in  sq.  in.  of  plates  on  one  edge  of  web 

plate  at  middle  of  span. 
d  =  distance  center  to  center  of  gravity  of  cover 

plates. 
/  =  unit  fiber  stress. 

The  total  moment  of  resistance  of  the  plate  girder  is  the  sum 
of  the  moments  of  resistance  of  the  web,  the  angles,  and  the  cover 
plates. 

The  rivets  used  to  connect  the  flange  angles  to  the  web  and  to 
connect  the  cover  plates  to  the  angles  must  be  spaced  to  take 


GIRDERS  AND  TRUSSES  115 

care  of  the  shear,  this  being  accomplished  by  using  the  following 
formula : 

rd 

p  =  r 

in  which  V  =  total  vertical  shear  at  the  section  considered, 
r  =  the  resistance  of  one  rivet, 
d  =  distance  in  inches  between  the  center  of  the  upper 

row  of  rivets  and  the  lower  row  of  rivets, 
p  =  pitch,  center  to  center  of  rivets,  hi  the  flange. 
The  bending  moment  due  to  uniform  load  varies  as  a  parabola, 
and  as  plate  girders  are  generally  designed  for  a  uniform  load  the 
cover  plates  are  varied  in  length  to  provide  enough  area  for  ten- 
sion or  compression.    They  extend  a  short  distance  past  the  point 
where  they  are  no  longer  required,  to  allow  for  proper  connec- 
tions.   By  having  the  plates  stop  when  no  longer  required  some 
weight  is  saved  and  the  design  is  the  most  economical  possible. 
When  concentrated  loads  must  be  cared  for  in  addition  to  a  uniform 
load  the  process  is  altered. 
Let  A  =  total  area  of  angles  and  cover  plates  in  one  flange  at 

mid-span, 

a\  =  area  of  shortest  cover  plate, 
a2  =  area  of  second  cover  plate, 

ax  =  area  of  longest  plate,  the  plates  being  numbered  pro- 
gressively from  1  to  the  end,  x  being  used  to  signify 
any  general  terminating  number. 

Similarly  li,  lz  •  •  •  In,  etc.  =  lengths  of  the  plates,  the  letter 
n  being  used  to  designate  the  general  number  applying  to  the  last 
plate,  thus  ln  =  length  of  the  last  plate  considered. 

L  

then,  In  =  -/=  X  V«i  +  «2  +  .  .  .  ax. 

The  angles  always  extend  the  full  length  of  the  web  plate. 

In  spacing  rivets  in  the  flanges  of  plate  girders  it  is  common 
practice  to  have  the  rivet  spacing  uniform  between  stiffeners,  the 
amount  of  vertical  shear  considered  as  being  taken  at  each  stif- 
fener.  The  minimum  distance  between  centers  of  rivets  is  three 
diameters  of  the  rivet,  but  not  less  than  3  ins.  for  f-in.  rivets  and 
2^  ins.  for  f-in.  rivets.  In  the  flanges  the  maximum  pitch  should 
not  exceed  six  times  the  diameter  of  the  rivets.  The  maximum 
pitch  at  ends  of  cover  plates  should  not  exceed  four  diameters  of 
the  rivets  for  a  length  equal  to  twice  the  depth  of  the  girder.  The 


116 


PRACTICAL  STRUCTURAL  DESIGN 


maximum  pitch  in  stiffeners  is  determined  by  the  loading,  if  any, 
but  should  never  exceed  a  maximum  of  4|  ins.  The  load  on  an 
intermediate  stiffener,  and  the  load  plus  the  end  shear  trans- 
mitted to  an  end  stiffener,  is  divided  by  the  bearing,  or  shearing, 
value  of  the  rivet  on  the  plate  to  obtain  the  number  of  rivets. 
These  are  equally  spaced,  but  the  maximum  spacing  is,  as  stated 
above,  4^  ins. 

When  it  is  necessary  to  splice  the  web  of  a  girder  the  splice 
plates  on  each  side  of  the  web  must  be  proportioned  so  the  rivets 
will  not  be  unduly  stressed.  The  moment  causing  vertical  bending 

in  the  girder  makes  the 
stress  in  the  rivets 
greater  as  the  bottom 
or  top  is  approached. 
Sometimes  an  addi- 
tional "  moment  splice  " 
must  be  added  on  the 
sides  of  the  web  close  to 
the  flange  angles. 

The  lengths  of  flange 
cover  plates  and  the 
spacing  of  rivets  may 
be  done  graphically. 
In  fact  the  graphical 
method  for  obtaining 
the  length  of  cover 
plates  is  that  commonly 


Fig.  73 


used  by  designers.  Fig.  29  shows  a  moment  diagram  on  a  beam 
carrying  a  uniform  load  and  several  distributed  loads.  Fig.  73 
shows  the  combined  curve  for  this  condition,  so  that  it  will  not 
be  necessary  to  measure  up  and  down  from  the  neutral  axis  and 
add  the  lengths. 

The  moment  curve  for  a  plate  girder  is  similarly  drawn.  For 
a  uniform  load  only,  the  curve  is  a  parabola,  so  it  is  necessary  to 
show  but  one-half  the  span.  All  horizontal  measurements  are 
made  to  the  scale  of  the  drawing,  and  all  vertical  measurements 
are  made  to  the  scale  used  for  the  bending  moment.  Make  the 
drawing  no  larger  than  is  necessary  to  obtain  accurate  data. 

To  set  off  the  lengths  of  cover  plates  draw  a  vertical  line  through 
the  point  of  maximum  bending  moment,  Fig.  74.  Begin  at  the 


GIRDERS  AND  TRUSSES 


117 


Cover  Plate 


Cover  Plate 


\ 


Fld'hge  bnjfgifr 


bottom  and  set  off  first  the  amount  of  moment  carried  by  the 
web  (|,  TV  or  TV,  according  to  specifications)  and  above  this  set 
off  the  amount  carried  by  the  flange  angles.  The  number  of  cover 
plates  having  been  determined,  set  off  in  succession  the  amount 
of  moment  carried  by  each.  Through  the  points  fixing  the  amount 
of  moment  carried  by  the  web  and  the  angles  draw  horizontal 

lines  to  the  ends  of , 

the  span.  Through 
the  points  fixing  the 
amount  of  moment 
carried  by  each  cover 
plate  draw  horizontal 
lines  beyond  the  mo- 
ment curve  2  or  3 
feet.  This  projection 
allows  length  in  which 
to  place  a  few  rivets 
so  the  plates  begin 
to  be  effective  when 
needed. 

The  lengths  of  the 
cover  plates  are  scaled 
from  the  diagram. 
When  the  plate  girder  Fig-  74 

carries  a  moving  load  on  top  it  is  usual  to  have  the  top  cover 
plate  next  the  angles  extend  the  full  length  of  the  girder.  This 
protects  the  angles  against  the  entry  of  moisture  in  the  joints  and 
stiffens  them  near  the  ends  against  the  effects  of  deflection  of  the 
frame  carrying  the  load. 

A  similar  diagram  may  be  used  for  spacing  rivets.  The  vertical 
scale  represents  the  total  maximum  tensile  (or  compressive)  stress 
in  the  girder,  instead  of  the  maximum  moment.  The  number 
of  rivets  necessary  to  resist  this  stress  is  determined  by  dividing 
the  stress  by  the  safe  allowable  stress  on  each  rivet.  The  vertical 
line  is  divided  into  as  many  parts  as  there  are  rivets  required. 
Through  the  division  points  draw  horizontal  lines,  Fig.  75,  to  an 
intersection  with  the  boundary  curve.  From  the  points  of  inter- 
section with  the  boundary  curve  drop  vertical  lines  to  the  base. 
A  rivet  will  be  placed  at  each  intersection  thus  determined.  A 
similar  method  may  be  used  for  spacing  stirrups  in  beams  of 


Plate  Girder       *  Heuiral  Axis, 


118 


PRACTICAL  STRUCTURAL  DESIGN 


reinforced  concrete.  The  moment  diagram  for  obtaining  the 
lengths  of  cover  plates  may  be  used  without  change  for  determin- 
ing rivet  spacing  by  adopting  a  scale  for  the  vertical  lines  pro- 
portioned to  the  ratio  the  total  stress  bears  to  the  moment.  By 
total  stress  is  meant  the  product  of  the  area  of  the  flange  angles 

plus  the  area  of  the 
cover  plates  multi- 
plied by  the  unit 
stress. 

Beams  with  Uniform 

Stress 

When  the  shape  of 
a  beam  resembles  the 
shape  of  the  bending 
moment  diagram  the 
stress  is  the  same 
along  the  length. 
When  the  top  and 
bottom  of  a  beam  are 
parallel  the  stress  di- 
minishes toward  the 
ends.  Cast-iron 
beams,  therefore,  are 
generally  made  with  a  "  belly,"  for  the  material  can  be  distributed 
at  will  since  the  beam  is  cast  in  a  mold.  This  effects  some  saving 
in  the  cost  of  patterns  and  castings.  Plate  girders  are  sometimes 
made  in  this  form,  a  familiar  example  being  the  main  girders 
along  the  under  side  of  railway  cars.  The  stress  equals  the  mo- 
ment at  any  point  divided  by  the  depth  of  the  beam  at  that 
point. 

Trussed  Beams 

The  trussed  beam  shown  in  Fig.  76  is  the  most  simple  form  of 
truss.  One-half  of  a  uniformly  distributed  load  is  assumed  to  be 
concentrated  over  the  strut  hi  the  middle,  so,  letting 


Fig.  75 


GIRDERS  AND  TRUSSES  119 

Referring  to  Figs.  48,  49,  50,  wherein  the  stress  is  shown  to 
be  equal  to  the  moment  divided  by  the  depth,  the  compressive 
stress  in  the  long  horizontal  member  is, 


The  length  of  the  diagonal  portion  of  the  tie  is  found  by  the 
formula  ,-j^ — ,n  j\2 

2 
and  the  tensile  stress  in  the  diagonal  is 

T  =  2d 

The  compressive  stress  in  the  vertical  strut  depends  upon  the 
construction  of  the  horizontal  member.    If  it  is  in  two  pieces  joined 
over  the  strut,  one- 
half  the  load,  P,  is    I  -~F~      -*f*-         \ ->| 

carried  by  the  strut. 

If  it  is  in  one  piece, 

or    composed    of 

several  planks  so  Fig  76. _ single  strut  Belly  Rod  Beam 

joined  that  they  act 

as  one  piece,  the  strut  carries  f  P  when  P  is  a  single  concentrated 

load  or  $W  when  IT  is  a  uniformly  distributed  load.     This  value 

must  be  used  for  P  in  the  above  formulas. 

When  the  single  strutted  beam  carries  a  single  concentrated 
load  over  the  strut  the  latter  carries  the  whole  load,  plus  half  the 
weight  of  the  uniformly  distributed  load  of  the  beam.  The  ten- 
sile and  compressive  stresses  in  the  horizontal  and  diagonal  members 
are  found  as  explained  above. 

In  Fig.  77  is  shown  a  beam  with  struts  at  the  third  points. 

The  bending  moment  for  a  beam  carrying  two  equal  loads,  P, 
at  a  distance  =  L  -r-  3  from  each  end,  is 


The  horizontal  stress  C  =  TTT* 
3d 

Pt 

The  diagonal  stress,  T  =  —=- 

The  compressive  stress  in  each  strut  =  P. 


120  PRACTICAL  STRUCTURAL  DESIGN 

When  the  load  is  uniformly  distributed  |  is  carried  by  each  strut, 

W 

so  in  the  above  expressions  substitute  -5-  for  P. 

o 

When  the  horizontal  member  is  in  one  piece  and  uniformly 
loaded,  each  strut  carries  -^  of  W. 

The  above  formulas  are  true  for  any  depth  of  truss  of  the  single  or 
double  strut  kind.  The  truss  may  be  reversed  so  that  the  sloping 
members  are  above  and  the  long  horizontal  member  is  below. 
Then  the  lower  member  carries  tension  and  the  upper  members 
are  in  compression.  The  methods  of  computation  are  not  altered 

r  ........  *•  ......  T  ........  *-±  ........  fr  ......  *! 


"-VU  .....  -r 
Fig.  77.  —  Double  Strut  Belly  Rod  Beam 

except  that  the  vertical  pieces  are  ties  instead  of  struts.  With 
the  load  applied  at  the  upper  end  the  ties  carry  no  stress  from  the 
load  but  are  used  merely  to  maintain  the  horizontality  of  the 
lower  chord.  With  the  load  applied  at  the  lower  end  each  vertical 
carries  the  amount  it  would  carry  if  the  beam  were  inverted.  With 
a  double-tied  beam  the  tie  serves  to  hold  the  frame  together  in 
case  of  a  rolling  load,  or  a  load  applied  other  than  vertically,  in 
which  case  it  does  carry  stress.  Diagonal  counters  set  between 
the  ties  will  take  care  of  such  stresses  and  the  ties  merely  serve 
to  hold  the  frame  together.  It  is  advisable  to  have  ties  many 
times  in  trusses  when  an  analysis  shows  they  are  not  stressed, 
in  order  to  carry  the  weight  of  the  lower  chord.  If  the  lower 
chord  must  carry  all  of  its  own  weight,  or  any  load  between  sup- 
ports, bending  and  shearing  stresses  will  be  set  up  in  the  lower 
chord  in  addition  to  the  direct  tensile  stress.  This  is  one  reason 
for  making  all  tension  members  of  metal  when  possible. 

Dimensions  are  on  center  lines.  The  tension  rods  should  go 
through  the  ends  of  the  compression  member  at  the  neutral  axis. 
The  plates  at  the  ends  should  be  normal  to  the  direption  of  the 
tie.  The  area  of  each  plate  is  obtained  by  dividing  the  tension 
in  the  rod  by  the  allowable  safe  unit  compressive  stress  on  the  end 
of  the  wood. 

The  area  of  each  strut  is  obtained  by  dividing  the  compression 


GIRDERS  AND  TRUSSES  121 

in  the  strut  by  the  safe  unit  compressive  stress  in  the  material 
of  which  it  is  made.  The  area  of  the  end  of  the  strut  against 
the  wood  is  found  by  dividing  the  compression  in  the  struts  by 
the  allowable  safe  unit  compressive  stress  across  the  grain  of  the 
wood. 

The  size  of  the  long  compression  member  is  obtained  by  design- 
ing it  as  a  column,  plus  the  effect  of  bending  caused  by  whatever 
load  it  may  carry  as  a  beam,  with  spans  figured  between  end  sup- 
ports and  vertical  struts,  or  ties.  The  unit  compressive  stress  on 
the  end  of  wood  may  be  used  when  the  length  of  the  member 
between  supports  does  not  exceed  15  times  the  least  thickness. 
For  longer  pieces  the  unit  compressive  stress  must  be  reduced  by 
an  appropriate  column  formula. 

The  sizes  of  metal  tension  members  are  fixed  by  dividing  the 
total  tension  by  the  allowable  safe  unit  tensile  stress  in  the  metal 
used.  If  threads  are  cut  in  a  rod,  this  size  must  be  at  the  root  of 
the  threads.  If  the  rod  has  upset  threads  the  full  area  of  the  rod 
may  be  used.  The  minimum  size  rod  to  use  in  any  tie  is  f  in. 
diameter. 

The  size  of  a  tension  member  made  of  wood  is  obtained  by 
dividing  the  total  tension  by  the  allowable  safe  tensile  stress  in 
the  wood  and  adding  thereto  an  area  equal  to  that  caused  by 
bolt  holes  and  seating  of  other  truss  members. 

Trusses 

A  truss  is  a  system  of  framework  forming  a  skeleton  beam.  The 
top  chord  is  in  compression;  the  bottom  chord  is  in  tension; 
the  web  members  (interior  braces  and  ties)  carry  the  shear.  The 
parts  must  be  in  equilibrium,  that  is  each  push  must  be  balanced 
by  a  pull  or  a  push  from  the  opposite  direction.  This  indicates 
the  triangle  as  the  perfect  truss,  for  it  cannot  be  changed  in  shape 
without  breaking  at  the  joints. 

Trusses  are  of  two  kinds,  —  those  with  parallel  chords  and  those 
with  nonparallel  chords.  The  parallel  chord  truss  will  first  be 
considered.  Fig.  78  shows  the  development  of  the  Pratt  truss 
from  two  panels  to  six  panels.  Assume  a  load  =  1  at  the  middle 
vertical.  Half  the  load  goes  to  each  support  so  the  coefficient 
for  the  two  middle  diagonal  ties  =  \.  If  the  load  is  applied  at 
the  top,  1  is  the  coefficient  for  the  middle  vertical,  but  if  the 
load  is  suspended  at  d  the  coefficient  =  0. 


122 


PRACTICAL  STRUCTURAL  DESIGN 


In  Fig.  78  (6)  two  panels  have  been  added  and  it  is  assumed 
the  panels  are  loaded  equally,  the  loads  being  concentrated  at 
the  joints.  The  end  panels  not  only  carry  the  reactions  from  the 
original  middle  triangle  but  their  own  loads  in  addition.  The 
diagonal  ties  ef  are  more  heavily  stressed  on  this  account  than  are 
the  ties  cd,  the  stress  being  tension,  for  the  whole  load  is  sus- 
pended at  the  end  points  /.  This  increased  tension  throws  on  the 


h        £         f 


b       9.       c 


f       s      A 


Load  on  Bottom  Chord 


Load  on  Top  Chord 


Fig.  78 


middle  of  the  upper  chord  the  added  load  of  the  end  panels  in 
addition  to  the  load  carried  by  the  middle  triangle,  so  the  middle 
panels  of  both  upper  and  lower  chords  are  more  heavily  stressed 
than  the  end  panels. 

The  coefficient  for  ad  and  cd  =  |.  When  the  unit  load,  1,  at 
the  next  joint  is  at  c  the  coefficient  for  ce  =  |  +  1  =  f ,  but  if  the 
load  is  at  e  the  coefficient  for  ce  =  \.  The  coefficient  for  ef  is  the 
sum  of  the  coefficient  for  dc  and  the  unit  load  on  the  line  ce.  The 
above  explanations  are  based  on  the  load  being  on  one  chord  only. 
When  there  are  loads  on  both  chords  there  should  be  two  sets  of 
coefficients  written  down  and  their  sum  used.  Using  the  positive 
sign  (+)  to  indicate  compression  and  the  negative  sign  (— )  to  indi- 
cate tension,  the  algebraic  sum  is  meant  when  the  stresses  are 
opposite  in  kind.  To  check  what  has  been  stated,  note  that  there 
are  three  panel  joints  carrying  equal  loads,  so  f  of  the  total  load 


GIRDERS  AND  TRUSSES  123 

goes  to  each  support.  If  the  load  is  uniformly  distributed  one- 
half  a  panel  load  will  be  concentrated  at/,  but  this  is  carried 
directly  on  the  supports  and  has  no  effect  on  the  stresses  in  the 
framework. 

In  Fig.  78  (c)  two  more  panels  have  been  added.  If  the  load  is 
on  the  top  chord  the  coefficients  are  as  follows: 

M  =  1  ce  =  f  fg  =  f 

cd  =  \  e/  =  f  gh  =  % 

If  the  load  is  on  the  bottom  chord : 

bd  =  0  ae  =  |  fg  =  f 

ad=%  <5f=f  0ft -f 

Merely  for  illustration  the  end  panels  have  been  completed  by 
dotted  lines.  The  coefficient  for  hi  =  f  (that  is,  it  carries  the  reac- 
tion). The  difference,  f  —  f  =  \  at  h,  acts  vertically  and  creates 
no  stress  in  the  truss.  The  member  gi  carries  no  load  when  the 
weights  all  act  vertically,  but  in  case  of  wind  or  rolling  loads  caus- 
ing horizontal  or  diagonal  action  on  the  frame  there  will  be  com- 
pression on  the  member  gi  at  the  end  where  the  load  is  applied 
and  tension  in  the  same  member  at  the  opposite  end.  When  the 
vertical  post  hi  is  omitted,  the  end  h  rests  on  the  abutment  and 
the  truss  is  said  to  be  suspended. 

Fig.  79  shows  the  development  of  the  Howe  truss,  which  is  merely 
the  Pratt  truss  inverted.  The  verticals  are  in  tension  and  the 
diagonals  are  in  compression.  The  Pratt  truss  is  usually  the  more 
economical  and  may  be  built  of  metal,  or  of  metal  and  wood.  The 
Howe  truss  is  usually  a  combination  of  metal  for  tension  members 
and  wood  for  compression  members.  For  maximum  economy  in 
metal  trusses  the  compression  members  should  be  as  short  as 
possible,  so  the  Howe  truss  is  not  well  adapted  for  all  metal 
construction. 

Coefficients  for  the  Howe  truss  are  written  as  explained  for 
the  Pratt  truss,  with  the  stresses  reversed  in  kind.  The  middle 
vertical,  however,  is  opposite  in  character  as  affected  by  the  load. 
That  is,  when  the  load  is  on  the  lower  chord  the  coefficient  =  1, 
but  when  it  is  on  the  upper  chord  the  coefficient  =  0.  Practically, 
however,  in  the  latter  case  the  vertical  does  carry  a  portion  of  the 
weight  of  the  lower  chord  in -the  middle  panel.  This  is  very  small 
and  the  smallest  sized  rod  used  will  more  than  take  care  of  it. 

A  coefficient  represents  the  proportion  of  panel  load  carried 
by  the  member  on  which  it  is  written.  Coefficients  are  used  only 


124 


PRACTICAL  STRUCTURAL  DESIGN 


when  all  the  panels  carry  equal  loads,  the  truss  then  being  sym- 
metrically loaded.  Instead  of  starting  from  the  middle  panel 
and  working  to  the  ends,  the  coefficients  may  be  obtained  as 
follows:  Count  each  panel  load  =  1.  One-half  the  number  of 
panel  loads  will  be  the  reaction  (expressed  as  fractional  coefficients). 
From  one  end  reaction  subtract  1  successively  at  each  panel 
joint  and  thus  obtain  the  coefficient  for  each  member  in  the 


(a) 


Fig.  79 

following  panel.  The  coefficient  for  the  chord  in  any  panel  is  the 
sum  of  the  coefficients  of  the  diagonals  between  that  panel  and  the 
end  support. 

The  weight  (TF)  carried  by  each  truss  member  is  equal  to  the 
coefficient  of  the  member  multiplied  by  the  unit  panel  load,  P. 

Let  I  =  length  of  panel,  center  to  center  of  verticals. 
d  =  depth  of  truss,  center  to  center  of  chords. 
t  =  length  of  diagonal,  center  to  center  of  chords, 
then    t  =  VI2  +  d*. 

Stress  in  diagonals  =  — r- 

Wl 

Stress  in  chords  =  —r' 
d 

Stress  in  verticals  =  W 


GIRDERS  AND  TRUSSES  125 

The  theory  of  coefficients  is  as  follows:  Assuming  the  truss 
to  be  symmetrically  loaded  with  uniform  loading  on  each  panel, 
half  the  load  on  the  middle  panel  alternately  pulls  and  pushes 
(or  pushes  and  pulls)  on  all  the  web  members  until  the  end  of  the 
truss  is  reached.  Each  panel  load,  as'  its  point  of  application  is 
reached,  is  added  and  the  end  web  member  carries  half  the  entire 
load  on  the  truss.  Thus  the  load  on  the  web  members  increases 
from  the  center  to  the  ends  and  the  load  on  a  chord  increases  from 
the  ends  to  the  center.  The  end  half  panel  load  is  carried  by  the 
abutment.  It  creates  no  stress  in  the  truss. 

For  irregular  and  unsymmetrical  loading  find  the  reactions  as 
for  a  simple  beam  similarly  loaded  with  concentrated  loads,  and 
the  panel  joint  where  the  maximum  moment  occurs  is  the  point 
of  zero  shear.  From  this  point  the  loads  run  up  and  down  the 
web  members  to  the  ends,  instead  of  from  the  center  panel,  as  in 
the  case  of  uniform  and  symmetrical  loading.  For  unsymmetri- 
cally  loaded  trusses  the  weight  per  panel  is  used  instead  of  the 
proportion  of  weight  (coefficient). 

A  truss  being  merely  a  skeleton  beam,  a  study  of  the  manner 
in  which  the  loads  go  to  the  abutments  shows  that  the  weight 
on  each  panel  is  really  the  shear  on  the  panel.  It  is  thus  feasible 
and  practical  to  consider  the  truss  as  a  beam  and  from  the  re- 
action at  either  end  subtract  in  succession  the  loads  on  the  panel 
joints  until  the  point  of  zero  shear  is  reached.  In  Fig.  80  is 
shown  a  truss  with  the  shear  diagram.  The  shear  on  gh  = 
25,000  Ibs. ;  on  ef  =  15,000  Ibs. ;  on  cd  =  5000  Ibs.,  the  panel  load 
being  10,000  Ibs.  concentrated  at  the  joints.  The  skeleton  truss 
lies  on  the  center  lines  of  the  members,  the  panel  length  being 
10  ft.  and  the  height  10  ft.  The  length  of  a  diagonal  =  14.14  ft., 

,       14.14  x  25,000 
so   the    compression  in   gh  = -^ — =  35,950  Ibs.     The 

,      14.14  x  15,000      01  __.    lu 
compression    in    ef  =  —    — -^ — •  —  =  21,750    Ibs.      The    com- 

14.  14.  v  ^000 

pression  in  dc  =  -      ^      -  =  7190  Ibs. 

With  the  load  considered  as  applied  on  the  upper  chord  the 
tension  in  gf  =  15,000  Ibs.;  the  tension  in  ec  =  5000  Ibs.;  and 
the  tension  in  bd  =  0.  With  the  load  considered  as  applied  on 
the  lower  chord  the  tension  in  gf  =  25,000  Ibs. ;  the  tension  in  ec 
=  15,000  Ibs.;  and  the  tension  in  bd  =  5000  Ibs. 


126 


PRACTICAL  STRUCTURAL  DESIGN 


Wl 

The  compression  and  tension  per  panel  in  the  chords  =  —,-> 

U 

therefore  compression  m  eg  =  tension  in  fh  =  25,000  Ibs.,  for  the 

Z      10 

ratio  1  =  77j  =  1-       ^ne    compression    m    de  =  tension   in    cf  = 

25,000  +  15,000  =  40,000  Ibs.   The  tension  in  be  =  25,000  +  15,000 
+  5000  =  45,000  Ibs. 
The  object  of  the  computations  being  to  obtain  the  stresses  so 


Fig.  80. 

the  members  may  be  proportioned,  the  method  above  given  of 
following  the  loads  from  joint  to  joint  and  obtaining  the  coefficients 
for  uniformly  and  symmetrically  loaded  beams,  or  of  obtaining 
the  shear  at  panel  joints  for  unsymmetrically  loaded  beams,  is 
adequate  and  simple. 

It  can  be  proven  that  a  truss  is  merely  a  skeleton  beam  by 
finding  the  shear  and  bending  moment  at  each  joint  and  then  divid- 
ing the  bending  moment  by  the  depth  obtain  the  stresses  in  the 
chords,  the  web  members  carrying  the  shear.  In  Fig.  80  the  end 
reactions  each  equal  25,000  Ibs.  Then 

M,  at/  =  10  x  25,000  =  250,000  ft.  Ibs. 

M,  at  g  =  0,  for  the  top  chord  rests  on  gh. 

M,  at  c  =  (20  x  25,000)  -  (10  x  10,000)  =  40,000  ft.  Ibs. 

M,  at  e  =  10  x  25,000  =  250,000  ft.  Ibs. 


GIRDERS  AND  TRUSSES 


127 


M,at6  =  (30x25,000) -(10x10,000+20x10,000)  =450,000  ft.  Ibs. 
M,  atd  =  (20  x  25,000)  -  (10  x  10,000)  =  400,000  ft.  Ibs. 

Dividing  the  moments  by  the  depth, 

250  000 
tension  in  fh  =  compression  in  eg  =  — —  =  25,000  Ibs., 


tension  in  cf  =  compression  in  de  = 


400,000 
10 


40,000  Ibs., 


.      .    ,        450,000       .  _  nAn  1U 
tension  in  be  =  —    —  =  45,000  Ibs., 


Fig.  81  shows  a  truss  having  an  odd  number  of  panels.    There 
is  no  stress  in  the  dotted  cross  diagonals  in  the  middle  panel  except 


Fig.  81. 

in  case  of  wind  or  rolling  loads,  or  otherwise  unbalanced  loading. 
Coefficients  may  readily  be  written  for  uniform  and  symmetrical 
loadings  for  this  case,  or  the  loads  may  be  followed  from  the  point 
of  zero  shear  in  cases  of  unsymmetrical  loading,  or  the  shear  method 
may  be  followed. 

In  Fig.  82  (c)  is  shown  a  truss  with  a  subvertical  and  sub- 
diagonal  at  each  end.  Such  an  arrangement  involves  the  con- 
sideration of  an  additional  triangle  in  which  half  the  weight  is 
added  to  the  load  at  6  and  is  then  carried  to  a,  the  other  half  being 
added  to  the  load  at  c.  This  arrangement  offers  no  difficulty  when 
figured  by  the  shear  method,  but  sometimes  causes  trouble  and 


128 


PRACTICAL  STRUCTURAL  DESIGN 


confusion  when  an  attempt  is  made  to  trace  out  the  loads  from 
the  middle  panel,  or  point  of  zero  shear. 

Some  trusses  have  nonparallel  chords.  The  shapes  vary  from 
those  higher  at  one  end,  as  in  Fig.  82  (a),  to  those  approaching  an 
arch  form  as  at  (6).  Part  of  the  shear  is  carried  by  the  sloping 
chord.  When  the  chord  stress  is  found  by  one  of  the  preceding 
methods  it  is  the  horizontal  stress.  For  a  sloping  chord  the  hori- 


zontal stress  must  be  multiplied  by  the  inclined  length  and  the 
product  divided  by  the  panel  length,  the  result  being  the  axial 
(longitudinal)  stress  in  the  inclined  member. 

In  the  Warren  truss  (Fig.  83)  the  stresses  in  the  web  members 
are  alternately  tension  and  compression,  the  light  lines  indicating 
tension  and  the  heavy  lines  compression.  Each  panel  is  an  equi- 
lateral triangle  and  in  the  figure  the  truss  is  a  single  system.  By 
using  another  set  of  triangles  and  placing  the  trusses  side  by  side 
so  one  triangle  overlaps  another  by  half  the  width,  we  obtain  a 
double  system.  Similarly,  we  may  use  a  triple-system  or  a  four- 
system  truss.  When  two  or  more  systems  are  used  the  result  is 
a  Latticed  Truss,  (Fig.  84). 

Let  W  =  total  load  on  the  truss,  uniformly  distributed, 
P  =  load  on  each  triangle, 
n  =  number  of  triangles  in  the  primary  single  system. 


GIRDERS  AND  TRUSSES 


129 


W 

Then,  in  a  single  system  truss,  P  =  — 


In  a  double-system  truss,  P  = 


ir 


(2n)  -  1 

W 

In  a  triple-system  truss,  P  =  ^-^ -• 

(6n)  —  i 

W 

In  a  four-system  truss,  P  =  -rr—r -• 

Having  found  the  panel  load,  P,  each  system  is  figured  as  a 
frame,  and  the  combined  strength  of  the  systems  determines  the 


(4      (?)      4     (I)       tt      (I)       4 

'X    I    A    I     A    4    /^ 


~<«       (f)  ^        (D  ^          fj" 

Fig.  83  —  Warren  Truss. 

strength  of  the  completed  truss.  The  systems  are  connected 
together  at  every  joint  where  the  members  meet  or  cross.  The 
lower  apices  are  the  panel  joints  when  the  load  is  on  the  lower 
chord  and  the  upper  apices  are  the  panel  joints  when  the  load  is 
on  the  upper  chord. 

In  Fig.  83  (a)  the  load  is  on  the  lower  chord  and  in  (6)  the  load 
is  on  the  upper  chord.  The  truss  is  here  assumed  to  be  uniformly 
and  symmetrically  loaded.  Coefficients  may  be  written  by  start- 
ing from  the  center  line  of  the  span.  On  the  upper  and  lower 
chords  are  placed  the  summation  of  the  coefficients  for  the  chords 
in  the  respective  panels.  In  Fig.  83  (a)  the  coefficient  for  L0Z/i  is 
one-half  that  for  U\  Uz  and  in  (6)  the  coefficient  for  Uo  U\  is  one- 
half  that  for  LiZ/2.  The  dotted  parallelograms  at  L0  and  Li, 
and  U0  and  Ui,  represent  to  scale  the  panel  load  set  off  vertically 


130  PRACTICAL  STRUCTURAL  DESIGN 

with  the  parallelogram  of  forces  completed  by  drawing  horizontal 
lines  to  intersect  the' triangles.  Then  a  =  stress  on  U\  Uz  =  stress 
on  L\Lz.  It  is  twice  6  which  represents  the  stress  on  Uo  U\  and 
Z/oLi.  The  thrust  of  the  brace  t/iL0  =  the  pull  of  the  tie  U0Li. 
It  is  resolved  at  the  point  of  support  on  the  abutment  into  a  hori- 
zontal component  along  the  chord,  and  a  downward  vertical  com- 
ponent, which  latter  is  resisted  by  the  upward  reaction  of  the 
abutment. 

A  usual  ratio  of  depth  to  span  in  trusses  is  one-tenth,  but 
circumstances  may  alter  this.  It  may  be  used  in  the  absence  of 
computations  to  ascertain  the  economic  depth  and  economic 


Fig.  84.  —  Multiple  System  Warren  Truss.     (Lattice  Truss.) 

ratio  of  depth  to  span.  For  Howe  trusses  the  best  angle  for  the 
diagonals  is  45  degrees.  When  any  different  angle  which  indicates 
a  panel  length  greater  or  less  than  the  depth  is  adopted,  the 
Pratt  truss  is  better.  For  trusses  of  the  Warren  type  the  angle 
should  be  60  degrees. 

Deflection  is  usually  taken  care  of  by  making  the  horizontal 
panel  length  at  the  upper  end  £  inch  longer  than  the  horizontal 
panel  length  at  the  lower  end,  in  every  ten  feet  of  span.  This 
does  not  alter  the  lengths  of  the  verticals  but  does  alter  the  lengths 
of  diagonals  and  when  the  truss  is  in  place  the  bottom  chord  will 
be  cambered  upward.  Were  it  perfectly  straight  it  would  appear  to 
the  eye  to  sag.  The  amount  of  camber  in  inches  is  found  as  follows: 

d  =  depth  of  truss  in  inches. 

s  =  span  of  chord  in  inches. 

c  =  camber  in  inches. 
Sd 

C  =  T 

In  some  of  the  figures  of  trusses  the  spaces  are  lettered.  This 
is  the  system  introduced  by  Mr.  Bow  for  the  graphical  analysis 
of  frames.  The  member  is  indicated  by  the  letters  between  which 
it  lies.  In  addition  to  this  system  of  lettering  the  spaces  the 
joints  are  sometimes  numbered.  The  spaces  are  lettered  to  identify 
the  member  in  the  graphical  analysis  and  the  joints  are  numbered 
only  when  the  detail  drawing  of  the  joint  is  to  be  referred  to. 


GIRDERS  AND  TRUSSES  131 

Some  of  the  trusses  shown  have  the  joints  lettered  with  a  capital 
U  on  the  upper  chord  and  a  capital  L  on  the  lower  chord.  The 
subscript  figure  represents  the  number  of  the  joint  from  the  left 
end,  the  joint,  or  joints,  at  the  abutments  being  0.  In  the  draw- 
ings a  joint  is  referred  to  by  the  U  or  L  and  the  subscript  indi- 
cating the  number  of  the  joint.  A  member  is  identified  by  giving 
the  letter  and  subscript  number  of  the  joint  at  each  end  of  the  mem- 
ber. This  method  of  identifying  joints  and  members  is  common. 

Architects  and  designers  of  buildings  have  to  deal  with  the 
simpler  forms  of  trusses,  but  when  it  is  desirable  to  introduce  the 
maximum  economy  into  a  design,  that  truss  is  most  economical  in 
which  the  stresses  in  the  chords  are  constant  from  end  to  end.  This 
points  to  a  truss  having  the  general  outline  of  a  bowstring  girder. 
The  top  chord  should  be  straight  and  not  curved  between  joints. 
To  obtain  a  curved  outline  for  a  roof  it  is  easy  to  use  fillers  or  vary 
the  depths  of  the  purlins  resting  on  the  trusses.  For  an  exposed 
chord  where  the  polygonal  form  would  be  unsightly  the  expedient 
is  sometimes  adopted  of  curving  the  segments,  thereby  introducing 
bent  beams  with  arching  action.  This  should  never  be  done.  It 
is  better  to  use  a  false  curved  chord  in  segments  to  hide  the  short 
straight  pieces. 

The  stresses  in  the  top  and  bottom  chord  of  a  bowstring  truss 
are  found  with  sufficient  accuracy  by  assuming  the  truss  to  be 
uniformly  loaded.  The  moment  divided  by  the  depth  gives  the 
maximum  stress  at  the  center  of  the  top  chord  and  throughout  the 
lower  chord,  the  formula  being 


in  which  T  =  total  tension, 

C  =  total  compression, 
I  =  length  in  feet, 
w  =  uniform  load  per  lineal  foot, 
d  =  depth  in  feet  at  center  of  span. 

The  chord  assumes  some  of  the  functions  of  braces  as  the  ends 
are  approached,  where  the  inclination  of  the  chord  increases,  and 
the  compression  is  nearly  uniform  throughout  the  length.  The 
compression  at  any  point  distant  y  feet  from  the  center  is  given 
by  the  following  formula,  _ 


132 


PRACTICAL  STRUCTURAL  DESIGN 


The  stress  in  the  braces  increases  from  the  ends  to  the  center, 
as  in  the  case  of  the  Queen  truss,  and  may  be  figured  the  same 
way.  The  vertical  rods  at  the  joints  are  in  tension  and  the  braces 
are  in  compression.  The  center  panel  is  usually  as  wide  as  the 
height,  which  decreases  the  angles  at  which  the  braces  are  set 
as  they  approach  the  ends. 

Steeply  pitched  roofs  of  the  Howe  truss  type  may  be  figured 
by  the  method  of  coefficients  when  the  loads  are  uniform  and 
symmetrically  placed.  They  may  be  figured  by  the  cumulative 
load  method  or  by  the  shear  method  when  unsymmetrically  or 


Rf3P     R*4P  R2 

Fig.  85A.  —  (a)  King  Truss:  (6)  (c)  (d)  Queen  Trusses, 
irregularly  loaded.  In  Fig.  85A  the  truss  with  one  vertical  is 
a  King  truss.  When  the  vertical  is  a  post  it  is  a  King  Post  truss 
and  when  the  vertical  is  a  tie  it  is  a  King  Rod  truss.  At  (6)  and 
(c)  are  shown  Queen  trusses,  these  being  known  by  the  number 
of  panels  into  which  they  are  divided  and,  like  the  King  truss, 
being  Queen  Post  or  Queen  Rod  trusses,  as  the  verticals  may  be 
posts  or  rods. 

In  a  system  of  roof  framing  all  longitudinal  members  are  called 
purlins  and  all  members  extending  from  the  eaves  to  the  ridge 
are  rafters.  The  top  chord  of  a  truss  is  composed  of  rafters. 
Main  purlins  extend  from  truss  to  truss,  resting  on  the  joints  at 
the  upper  ends  of  verticals.  Intermediate  rafters  rest  on  the 
main  purlins  when  the  spacing  between  trusses  is  considerable 
and  across  the  rafters  sheathing  is  placed  to  carry  the  roof  cover- 
ing. By  so  doing  all  loads  are  concentrated  at  the  upper  ends  of 
the  verticals,  so  the  truss  rafters  (upper  chord)  are  in  compres- 


GIRDERS  AND  TRUSSES  133 

sion.  Sometimes  no  intermediate  rafters  are  used,  the  roofing 
being  carried  by  purlins  resting  on  the  joints. 

To  obtain  proper  results  the  sloping  rafter  of  the  truss  is  divided 
into  equal  spaces  and  verticals  are  dropped  to  the  bottom  chord 
(or  tie).  Braces  extend  from  the  foot  of  one  vertical  to  the  top  of 
another.  If,  through  any  error  or  because  it  is  considered  best, 
intermediate  purlins  rest  on  the  truss  rafters,  or  the  roof  is  carried 
directly  on  these  rafters,  it  will  be  necessary  to  design  them  to 
carry  the  bending  stress  in  addition  to  the  direct  compression. 

In  the  King  truss,  Fig.  85A  (a),  the  load,  P,  when  applied  at 
the  upper  vertex  causes  no  stress  in  the  rod  BB.  When  applied  at 
the  lower  end  the  stress  is  tension  and  equal  to  the  load.  The  stress 
AB  =  half  the  load,  so  the  coefficient  =  \.  The  stress  in  the 

horizontal  tie  rod  =  half  the  load  x  —  ,    .  ,     —  >  that  is 

height 


In  the  Queen  truss  the  action  resembles  an  arch  in  that  the  com- 
pressive  and  tensile  stresses  increase  towards  the  supports  in  the 
rafters  and  tie,  and  the  stresses  in  the  verticals  and  diagonals 
decrease  toward  the  supports,  for  the  inclined  rafters  carry  part  of 
the  shear.  Half  of  the  load  on  each  end  panel  is  carried  by  the 
abutments  and  creates  no  stress  in  the  truss. 

In  Fig.  85A  (6)  a  load  is  assumed  to  be  applied  at  the  upper  end 
of  BC.  If  the  load  is  at  the  lower  end  the  rod  BC  carries  this 
load  to  the  rafter  at  the  vertex  of  the  triangle.  If  the  load  is 
applied  directly  to  the  rafter  at  the  vertex  there  is  no  stress  in  BC. 
This  will  not  again  be  referred  to,  as  it  applies  to  the  rod  in  the 
end  triangle  in  all  trusses.  Half  the  load  is  carried  on  AB  and  half 
goes  down  CD  to  the  tie.  The  load  on  the  top  of  the  truss  is  in- 
creased by  the  load  coming  to  the  tie  by  the  braces  CD  on  both 
sides  of  the  center;  therefore  it  is  2P,  if  the  load  on  a  joint  is  called 
P.  The  vertical  rod,  DD,  however,  carries  only  the  load  at  the 
lower  end.  This  load  2P  from  the  top  of  the  truss  is  carried  half- 

way down  the  truss  to  the  joint  and  there  it  has  added  to  it  the 

p 
load  -~   of  the  brace  AB,  this  brace  therefore  carrying  a  load 

3P 
=  —  •     The  tie  rod  is  in  tension  by  an  amount 

3P      half  the  span          „      3P       L      3PL 
=  T  >         height         °r    1  =  ~2~  X  2d  =  ~ 


134  PRACTICAL  STRUCTURAL  DESIGN 

In  Fig.  85A  (c)  the  rafter  is  divided  into  three  equal  parts.    Each 

P  3P 

joint  carries  a  load,  P,    The  load  on  DC  =  -^ ;   on  DE  =  -^  ; 

KD 

on  EF  =  -§-    The  vertical  FF  carries  3P.    The  rafter  AF  carries 

3P  5P 

-jr- ;  AD  carries  -x-;   A  5  carries  3P.    The  tie  rod  carries  at  the 
z  & 

o  p  r 

end  a  stress  =  -^T"     The  stress  in  the  tie  rod  on  the  section 

K  p  j 
TE=       ,  •     In  actual  practice  the  tie  rod  is  uniform  in  size 

throughout  the  span. 

With  the  examples  given  the  student  should  have  no  trouble 
tracing  the  loads  on  the  members  of  the  truss  shown  at  (d). 

Each  vertical  is  hi  tension  by  an  amount  equal  to  the  load  it 
carries.  Each  diagonal  member  is  in  compression  by  an  amount 

=  ^7-     in  which  x  =  amount  of  load  on  the  member, 

L  =  length  of  member, 

d  =  the  vertical  height  from  the  bottom  to  the 
top  of  the  member. 

All  measurements  are  on  center  lines.  The  slope  of  the  rafter 
is  constant  so  the  ratio  is  obtained  once  by  dividing  the  slant 
length  of  the  rafter  by  the  height  of  the  truss.  The  slopes  change 
at  each  panel  for  the  interior  braces,  so  a  ratio  must  be  found  for 
each  separately. 

Coefficients  for  Fink  trusses,  Fan  trusses,  and  Pratt  trusses  with 
inclined  rafters  have  been  calculated  for  different  degrees  of  slope 
and  for  varying  numbers  of  panels,  based  on  uniform  symmetrical 
loads.  Tables  of  these  coefficients  are  given  on  pages  309-311,  of 
the  1913  edition  of  the  "  Carnegie  Pocket  Companion  "  for  trusses 
to  be  made  of  steel  or  wrought  iron.  Steel  is  commonly  used  except 
when  corrosion  is  a  grave  danger,  in  which  case  wrought  iron  is 
preferred.  All  metal  trusses  are  made  of  rolled  shapes  with 
riveted  connections.  The  trusses  illustrated  may  be  a  combina- 
tion of  steel  rods  for  tension  members  and  wood  for  compression 
members.  Fink  trusses  are  very  generally  used  because  most  of 
the  members  are  in  tension  and  the  struts  are  short.  Partial 
loading  can  never  cause  maximum  stresses  in  the  parts  of  Fink 
trusses  as  they  may  in  other  forms  of  trusses. 


GIRDERS  AND  TRUSSES  135 

Roof  Loads 

For  information  as  to  proper  roof  loads  and  the  effect  of  wind 
the  student  is  referred  to  pages  305-307,  1913  edition  "Carnegie 
Pocket  Companion."  This  will  also  be  dealt  with  in  the  chapter 
on  "  Graphic  Statics."  Usually  city  ordinances  specify  that  a 
roof  shall  be  capable  of  carrying  40  Ibs.  per  square  foot  of  hori- 
zontal surface,  in  addition  to  its  own  weight,. this  allowing  for 
wind,  snow,  live  load,  and  roofing.  Some  cities  require  only  25  Ibs. 
and  others  30  Ibs.  For  a  steeply  pitched  roof  25  Ibs.  is  proper,  but 
for  a  very  flat  roof  the  designing  load  should  not  be  less  than  50 
Ibs.  per  square  foot.  Each  joint  in  a  frame  carries  a  load,  P, 
equal  to  the  truss  spacing  times  the  panel  length  multiplied  by 
the  load  per  square  foot. 

The  Signs  Used  for  Stresses 

The  author  mentioned  that  the  positive  (+)  sign  indicates 
compression  and  the  negative  (— )  sign  indicates  tension.  This 
is  the  way  he  was  taught,  and  thirty  years  ago  this  use  of  the 
signs  was  common  with  American  and  British  writers.  There 
was  a  certain  mnemonic  aid  in  using  the  signs  thus,  for  compres- 
sion thickens  a  body  and  tension  makes  it  thinner,  so  the  "  minus 
sign  "  expressed  the  idea  of  thinness.  In  drawings  the  pieces 
in  compression  were  indicated  by  heavy  lines  and  the  pieces  in 
tension  by  light  lines. 

Continental  European  writers  used  the  signs  in  a  directly 
opposite  sense,  for  strict  mathematical  analysis  in  which  careful 
attention  must  be  paid  to  the  signs  of  quantities  resulted  in  bring- 
ing compression  out  at  the  end  with  a  negative  sign  and  tension 
with  a  positive  sign.  The  well-trained  mathematician  needs  no 
aid  from  mnemonics.  The  result  of  late  years  has  been  to  unsettle 
American  and  British  authors,  and  a  reader  of  modern  books  must 
be  careful  to  ascertain  just  how  the  signs  are  used  by  the  author. 
It  is  to  be  hoped  that  at  some  not  distant  day  all  writers  will 
agree  upon  a  definite  use  of  the  signs,  but  for  the  purposes  of  the 
present  work  the  author  believes  the  mnemonic  value,  as  given 
above,  is  too  great  to  be  neglected. 


85B- 


CHAPTER  V 
Joints  and  Connections 

FIG.  85s  was  copied  from  a  sheet  of  drawings  forming  part  of 
a  set  made  for  the  building  of  a  public  school  in  a  middle 
western  state.    The  writer  knows  from  experience  in  check- 
ing designs  that  this  is  a  common  type  of  roof.    Amateur  architects 
and  many  young  draftsmen  have  a  fondness  for  constructing 
wooden    trusses    with    light    tension    members    of  wood.    The 
student  who  read  Chapter  IV  carefully  knows  how  simple  a 
Usheefing  matter  it  is    to  find 

'  .composition***    ,/^'sfnp  rv a^fa.  ^       thestressesinthe 

members  of  a  simple 
truss.  There  are  many 
hand-books  on  the 
market  which  contain 
diagrams  and  formu- 
las for  the  design  of 
trusses,  so  it  is  not  a  difficult  thing  to  proportion  the  members 
of  any  form  of  truss. 

The  student  is  advised  to  check  the  design  of  the  truss  shown 
in  Fig.  85s.  It  will  be  a  very  useful  exercise.  The  joists  are  on 
16-in.  centers.  The  joists  under  the  roof  are  2  in.  x  8  in.  and 
carrying  sheathing  on  which  is  placed  the  composition  roofing. 
The  ceiling  joists  are  2  in.  x  6  in.  In  the  design  assume  a  fiber 
stress  hi  tension  and  compression  of  800  Ibs.  per  sq.  in.,  the  loads 
per  square  foot  of  horizontal  surface  to  be  as  follows:  composition 
roofing,  5  Ibs. ;  joists,  5  Ibs.;  sheathing,  4  Ibs.;  plastering,  5  Ibs.; 
roof  load  30  Ibs. ;  truss,  4  Ibs. 

In  the  specifications  appeared  the  following  clause:  "  The  truss 
is  to  be  well  and  securely  nailed  together,  but  the  number  of  nails 
at  each  joint  shall  not  be  less  than  shown  on  the  drawings."  The 
draftsman  had  placed  eight  dots  at  each  joint,  probably  meaning 
each  dot  to  represent  a  nail.  The  size  of  nail  to  use  was  not  given. 
On  the  advice  of  the  author  the  roof  was  changed  so  it  appeared 

136 


JOINTS   AND   CONNECTIONS  137 

as  if  inverted,  putting  all  the  diagonal  members  in  compression 
and  substituting  vertical  wrought  iron  rods  for  the  thin  boards 
used  in  the  original  design.  Before  accepting  the  change,  however, 
the  architect  attempted  to  retain  his  truss  by  substituting  in  each 
joint  four  f-in.  bolts,  saying  that  he  had  been  architect  for  more 
than  twenty  schoolhouses,  in  which  he  had  used  the  form  of  truss 
shown  in  his  drawings. 

When  a  member  is  in  compression  the  joint  is  not  so  hard  to 
construct,  or  design,  as  a  connection  at  the  end  of  a  tension  mem- 
ber. It  is  an  elemental  fact  that  when  a  piece  in  compression 
rests  on  another  piece  there  must  be  a  bearing  area  of  the  proper 
size  to  prevent  crushing.  It  is  only  in  joints  for  pieces  in  tension 
that  the  average  draftsman  seems  to  forget  elementary  principles. 
He  laps  one  piece  over  another,  specifies  that  the  contractor  shall 
"  nail  it  securely,"  and  passes  on  to  something  else.  It  is  either 
because  of  laziness  or  downright  ignorance  that  such  things  happen. 
Experienced  contractors  are  often  life  savers  for  incompetent 
draftsmen,  for  we  can  hardly  call  them  designers. 

When  there  is  a  pull  in  a  board  used  to  transmit  tension  in  a 
truss  the  joint  must  be  strong  enough  to  resist  the  pull.  It  is 
necessary  to  decide  on  the  size  of  nail  to  use  and  then  divide  the 
total  pull  by  the  resisting  power  of  one  nail  to  obtain  the  number 
of  nails  to  use.  To  avoid  splitting  the  wood  the  nails  should  be 
separated  by  a  space  at  least  twenty  times  the  thickness  of  the 
nail.  Even  this  rule  must  be  modified  by  the  sort  of  wood  used, 
as  some  wood  is  brittle  and  cannot  stand  many  nails  in  a  small 
space.  Wire  nails  do  not  cut  through  the  fibers,  but  spread  them 
apart  so  it  is  not  necessary  to  make  up  the  area  occupied  by  the 
nails,  but  this  must  be  considered  in  using  cut  nails.  The  proper 
size  of  nail  to  use  is  governed  largely  by  the  length.  It  must  be 
not  less  than  three  times  the  length  of  the  thinnest  outside  piece. 
If  it  cannot  go  two-thirds  its  length  into  wood,  because  the  wood 
is  not  thick  enough,  the  end  must  be  firmly  clinched. 

A  rule  often  blindly  given  in  pocket  books  for  the  lateral  resisting 
value  in  pounds  for  nails  is  as  follows: 
P  =  Cd,  in  which 

P  =  total  number  of  pounds  transverse  load  per  nail, 
C  =  a  coefficient  varying  from  4.5  to  12,  depending  on  the  nail, 

whether  wire  or  cut,  and  on  the  wood, 
d  =  the  size  of  nail  in  pennyweights. 


138  PRACTICAL  STRUCTURAL  DESIGN 

A  more  logical  rule  given  by  Mr.  H.  D.  Dewell,  in  Western 
Engineering,  Vol.  7,  page  291,  is  as  follows  for  Douglas  Fir, 

P  =  4000  d\ 

in  which  d  =  diameter  of  nail  in  inches. 

For  other  woods  multiply  the  result  by  the  following  coefficients: 

Long-leaf  yellow  pine 1.05 

White  pine 78 

Norway  pine 65 

White  oak -. 78 

For  common  wood  screws  use  the  constant  4375  instead  of  the 
constant  4000  used  for  nails. 

Nails  or  common  wood  screws  are  generally  thought  of  first 
for  fastening  timber  because  they  are  cheap  and  the  labor  cost  of 
driving  them  is  low.  Their  usefulness,  however,  is  limited  to  thin 
pieces  carrying  little  stress.  When  the  load  is  too  great  to  be 
transmitted  properly  by  nails  or  common  screws,  or  the  pieces 
are  too  thick,  lag  screws  may  be  used  on  account  of  the  low 
labor  cost  as  compared  with  that  required  for  bolts,  for  which 
holes  must  first  be  bored.  Mr.  H.  D.  Dewell,  in  Engineering 
News,  Vol.  76,  page  797,  gives  the  following  recommended  working 
values  for  lag  screws. 

Description  Pounds  per  screw 

Metal  plate  lagged  to  timber,  H  X  4^-in.  screw 1030 

Metal  plate  lagged  to  timber,  %  x  5-in.  screw 1200 

Timber  planking  lagged  to  timber,  %  x  4J^-in.  screw 900 

Timber  planking  lagged  to  timber,  J^  x  5-in.  screw 1050 

Generally  speaking  the  resistance  of  lag  screws  varies  with  the 
ratio  of  their  diameters,  so  the  values  above  given  may  be  used 
as  a  basis  for  other  sizes. 

The  strength  of  nails,  lag  screws,  and  bolts  in  wood  cannot  be 
computed  the  same  as  rivets  in  metal,  for  the  rivets  may  shear, 
but  this  is  impossible  with  joints  in  wood.  Nails,  screws,  or  bolts 
will  bend,  for  the  wood  will  crush  long  before  the  shearing  strength 
of  the  metal  is  reached.  It  is  necessary,  therefore,  to  use  bearing 
values  obtained  by  experiments.  Mr.  H.  D.  Dewell,  in  Engineer- 
ing News,  Vol.  76,  page  115,  described  the  result  of  tests  made 
with  bolts.  Two  thin  pieces  of  timber  were  fastened  to  a  thick 
piece,  by  bolts  passing  through  with  washers  on  the  ends.  The 


JOINTS   AND   CONNECTIONS 


139 


two  side  pieces  were  placed  on  the  table  of  a  testing  machine  and 
the  center  piece  was  pressed  down  until  failure  resulted. 

The  following  table  is  recommended  as  giving  proper  safe  loads 
corresponding  to  a  slip  not  exceeding  -^  in.  They  are  for  end 
bearing  with  bolts  having  a  driving  fit  and  the  thickness  of  each 
side  piece  equal  to,  or  greater  than,  one-half  the  thickness  of  the 
main  timber.  For  side  bearing  (across  the  grain)  the  values  can 
be  taken  at  six-tenths  the  values  given  for  end  bearing  in  the 
table. 

The  values  in  the  table  are  for  double  shear,  that  is,  for  three 
pieces  of  timber  having  two  shearing  planes.  For  cases  of  single 
shear,  two  timbers  bolted  together,  use  one-half  the  values  given 
in  the  table. 

The  working  values  given  are  for  Douglas  fir.  For  other  timbers 
the  values  are  to  be  multiplied  by  the  factors  following: 

Long-leaf  yellow  pine 1.05 

White  pine 78 

Norway  pine 65 

White  oak . .  .   .78 


TABLE  OF  WORKING  STRENGTH  OP  ONE  BOLT  IN  TIMBER  JOINT 
IN  DRY  TIMBER,  AS  FOR  USE  IN  INTERIOR  OF  BUILDING 

(Bolt  in  double  shear  bearing  against  end  of  grain.) 


Size  of  Bolt 
Ins.  diam. 


Thickness  of  One  Side  Piece  ^  One-half 
Thickness  of  Center  Timber  = 


2  Ins. 

3  Ins. 

4  Ins. 

6  Ins. 

5A 

1057 

1275 

1460 

1460 

% 

1450 

1665 

1980 

2100 

H 

1900 

2130 

2450 

2850 

i 

2460 

2664 

2970 

3705 

For  single  shear  take  one-half  the  above  values.  For  bearing 
against  the  side  of  grain  take  six-tenths  the  above  values. 

Joints  in  timber  are  sometimes  made  with  a  large  pin  on  which 
a  timber  rests.  The  pieces  are  joined  at  an  angle  less  than  90 
degrees  and  the  bearing  per  square  inch  on  the  fibers  will  be  some- 
thing less  than  the  allowable  safe  bearing  with  the  grain  and  con- 
siderably greater  than  the  allowable  safe  bearing  across  the  grain 
for  broad  surfaces. 


140 


PRACTICAL  STRUCTURAL  DESIGN 


n  = 


Let  n  =  the  bearing  per  sq.  in.  on  the  diametrical  area  of  the 

pin,  having  a  driving  fit. 

p  =  the  allowable  bearing  per  sq.  in.  on  the  ends  of  the  fibers, 
q  =  the  allowable  bearing  per  sq.  in.  perpendicular  to  the 
direction  of  the  fibers. 
+  f  ?•     (Dewell  formula) 

Example.  —  A  brace  rests  against  a 
2-inch  pipe  driven  through  the  bottom 
chord  of  a  truss  to  act  as  a  pin  joint. 
The  allowable  bearing  on  the  ends  of 
the  fibers  is  1200  Ibs.  per  square  inch 
and  the  allowable  cross  bearing  is  300 
Ibs.  per  square  inch. 

What  bearing  can  be  used  on  the 
diametral  area  of  the  pin? 

Answer.  —  (\  x  1200)  +  (f  x  300)  = 
Fig-  86-  600  Ibs.  per  sq.  in. 

Mr.  H.  D.  Dewell  was  Chief  Structural  Engineer,  Panama- 
Pacific  International  Exposition,  San  Francisco,  1915.  The 
possibilities  of  timber  construction  were  probably  never  better 
treated  than  in  the  work  under  his  charge,  and  a  great  many 
experiments  and  detailed  studies  were  made  in  order  to  design 
properly.  The  results  were  given  by  him  in  a  series  of  articles 
in  the  June  to  December  issues  (inclusive),  1916,  of  Western  Engi- 
neering, San  Francisco,  Cal.  The  author  has  made  free  use  of  the 
articles  by  Mr.  Dewell  and  discarded  much  material  prepared  for 
this  chapter,  based  on  older  writings, 
some  of  which  were  speculative  and 
some  of  which  were  founded  on  experi- 
mental work  conducted  by  men  not  so 
skilled  as  are  the  modern  experimenters. 
Standards  are  well  enough  in  their 
places,  but  men  should  not  blindly  use 
standards  without  knowing  all  reasons 
and  the  authority.  Standard  washers  should  not  be  used,  merely 
on  the  advice  of  an  advertisement  writer,  to  carry  certain  loads 
until  their  sufficiency  has  been  checked  by  computations  for  the 
wood  with  which  they  are  to  be  used.  If  a  washer  is  too  small  it 
will  be  drawn  into,  and  crush  the  fibers  of,  the  wood. 
Bolts  placed  through  the  joints  of  trusses  will  have  washers  as 


Fig.  87. 


JOINTS   AND   CONNECTIONS  141 

shown  in  Fig.  87.  One  washer  will  bear  on  the  side  of  the  fibers 
and  the  other  washer  will  bear  at  an  angle  to  the  fibers.  Using 
the  proper  table  giving  the  safe  bearing  strength  of  the  wood 
perpendicular  to  the  fibers  proportion  the  side  bearing  washer 
accordingly.  For  the  safe  pressure  to  allow  under  the  washer 
set  at  an  angle  to  the  fibers,  and  also  to  determine  the  safe  bearing 
at  the  foot  of  the  sloping  member,  several  formulas  have  been 
proposed. 
The  Jacoby  formula  is  as  follows: 

n  =  p  sin20  +  q  cos20, 

as  given  in  his  book  "  Structural  Details."  In  Engineering  News, 
Vol.  68,  Professor  Malverd  A.  Howe  published  the  result  of  some 
tests  made  to  determine  the  allowable  bearing  pressure  on  inclined 
surfaces  for  various  timbers,  and  recommended  the  formula 

n  =  q  +  (p  -  q)  (oQpJ 

in  which  n  =  the  allowable  unit  stress  on  a  surface  which  makes 

an  angle  6  with  the  direction  of  the  fibers, 
p  =  the  allowable  unit  stress  against  the  ends  of  the  fibers, 
q  =  the  allowable  unit  stress  on  the  sides  of  the  fibers. 
The  author  in  attempting  to  simplify  the  formula  of  Professor 
Jacoby  and  make  it  fit  closer  to  some  experiments  on  timber 
other  than  yellow  pine,  which  the  Jacoby  formula  closely  fitted 
at  low  angles,  developed  the  following  straight-line  formula. 

n  =  r^r>  minimum  value  equal  to  q.     Straight  line  80°  to  90°. 

Later,  learning  of  the  Howe  formula,  he  attempted  to  simplify 
it  and  developed  the  following  formula: 

t\        /    /)   \2 

n  =  77-57  f  77^7: )  >  minimum  value  equal  to  q. 
U.ol    \1UU/ 

In  the  two  formulas  of  the  author  the  angle  6  is  expressed  in 
figures,  as  10,  20,  30,  etc. 

In  Fig.  88  the  four  formulas  are  platted.  The  diagram  appeared 
in  the  July,  1916,  Western  Engineering,  with  the  Jacoby  and  Howe 
formulas,  the  author  adding  here  the  curves  produced  by  his  own 
formulas. 

The  student  should  now  be  able  to  determine  the  size  and 
number  of  nails,  screws,  lag  screws,  or  bolts  for  all  the  joints  of  the 


142 


PRACTICAL  STRUCTURAL  DESIGN 


truss  shown  in  Fig.  85s.  It  being  assumed  that  wire  nails  and 
common  screws  merely  push  the  fibers  aside,  no  additional  area  is 
required  on  account  of  the  holes.  Lag  screws  and  bolts,  however, 
cut  the  fibers,  so  it  is  necessary  to  add  to  the  members  a  width 
equal  to  the  diameters  of  the  holes.  By  the  time  the  student  has 

worked  through  the 
problem  of  detail- 
ing all  the  joints  he 
will  probably  learn 
that  there  is  not 
area  enough  for 
fastenings.  The 
problem  will  serve 
to  show  why  tim- 
ber should  not  be 
used  for  members 
in  tension  unless 
the  load  from  other 
It  is  best  when  using 
Use  wrought  iron  or 


20      30      40       50 
Value  of  Bin  Degrees. 

Fig.  88. 


70       60    "90 


members  can  be  transferred  by  end  thrust, 
wood  to  have  all  joints  in  compression, 
steel  for  tension  members. 

Fig.  89  illustrates  several  types  of  joints  and  fastenings  used  in 
framing  timber.  Bolts  in  carpentry  should  be  used,  when  pos- 
sible, only  to  hold  abutting  portions  of  timber  together.  It  is  not 
always  possible  to  so  use  them,  but  the  hint  is  enough  to  set  a  man 
to  studying  seriously  every  joint  he  makes  in  order  to  hold  to  the 
rule  if  possible.  Straps  are  more  expensive  than  bolts  and  are  not 
so  good.  A  detail  that  is  too  common  and  which  should  never 
be  used  is  shown  at  (a);  what  happens  when  the  wood  shrinks 
is  shown  at  (6),  for  it  is  impossible  to  tighten  the  joint.  Water 
getting  into  the  toe  of  a  joint  often  causes  the  fibers  to  decay  and 
this  added  to  shrinkage  ruins  the  truss.  A  preferable  method  is 
shown  at  (c).  The  bolt  may  be  tightened  from  time  to  time.  This 
form  of  joint,  however,  should  never  be  designed  with  part  of  the 
load  carried  by  the  abutting  end  and  part  by  the  inclined  bolt. 
Many  experiments  have  demonstrated  that  under  test  the  two  do 
not  act  together.  The  weaker  system  will  act  first  and  give  way 
before  the  other  gets  into  action.  It  is  best  to  design  for  the  entire 
load  to  be  taken  by  the  dap  and  use  the  inclined  bolt  only  for  the 
purpose  of  holding  the  pieces  in  contact.  All  bolts  will  be  screwed 


JOINTS   AND   CONNECTIONS 


143 


tight  so  that  washers  should  be  designed  with  area  enough  to  de- 
velop the  safe  working  stress  in  the  bolts  without  exceeding  the 
safe  bearing  on  the  wood.  If  the  safe  bearing  is  exceeded,  which 
will  be  the  case  if 
the  washers  are  too 
small,  the  fibers  will 
be  cut  around  the 
edges  of  the  washers 
and  decay  will  set  in. 
At  (d)  is  shown  a 
method  sometimes 
adopted  for  upper 
chord  joints  and  at 
(e)  is  shown  what 
happens  when  the 
wood  shrinks.  A 
proper  joint  is  shown 
at  ( / )  with  the  cen- 
ter lines  of  all  pieces 
meeting  at  a  com- 
mon point.  This 
idea  of  all  the  center 
lines  meeting  at  a 
common  point  is  also 
illustrated  at  (0)  and 
(ti).  This  is  neces- 
sary to  avoid  rota- 


Fig.  89 


tion  which  would  cause  a  bending  moment.  When  part  of  a  piece 
of  timber  is  cut  out  for  a  dap  the  stress  is  concentrated  in  the 
rest  of  the  piece,  which  alters  the  position  of  the  center  lines. 
This  must  be  considered  in  the  design  and  will  be  discussed  when 
we  take  up  the  design  of  joints. 

At  (i)  is  shown  a  brace  abutting  on  a  post.  If  a  waling  is  spiked 
along  a  line  of  posts,  or  blocks  with  the  fibers  in  a  horizontal  posi- 
tion are  bolted  to  the  posts,  and  the  braces  rest  on  the  sides  of  the 
fibers,  there  will  be  settlement  when  shrinkage  occurs.  When 
blocks,  or  cleats,  are  used  as  shown  they  should  have  the  fibers 
vertical  and  the  dap  should  be  deep  enough  to  transmit  all  the 
load  to  the  post.  The  bolts  shown  are  used  only  to  hold  the  cleats 
in  place.  To  design  the  bolts,  however,  it  is  well  to  have  them 


144  PRACTICAL  STRUCTURAL  DESIGN 

strong  enough  to  carry  half  the  load,  which  will  maintain  the 
integrity  of  the  joint  in  case  the  cleats  through  shrinkage,  or  decay 
of  dap,  lose  one-half  their  strength.  To  prevent  the  braces  from 
being  pushed  off  to  one  side  they  may  be  nailed  to  the  cleats  or 
a  thin  strip  of  iron  may  be  inserted  in  a  saw  cut,  one-half  in  the 
cleat  and  one-half  in  the  foot  of  the  brace. 

At  ( j)  is  shown  a  method  often  used  in  making  a  joint  at  the 
foot  of  a  raking  member.  The  vertical  bolts  are  assumed  to  resist 
the  thrust.  It  is  a  poor  joint.  The  habit  of  many  draftsmen  is 
to  guess  at  the  number  of  bolts,  and  they  seldom  add  additional 
area  to  supply  that  lost  in  the  bolt  holes.  A  better  joint  is  shown 
at  (&).  ^The  black  circles  are  the  ends  of  pipes  used  as  shear  pins. 
They  are  designed  to  take  all  the  thrust  and  the  joint  is  made  by 
spiking  the  two  pieces  together,  after  which  the  bolt  holes  are 
bored  and  the  bolts  driven  and  tightened.  Enough  bolts  should 
be  used  to  take  half  the  tension  in  the  lower  chord;  as  direct 
tension,  not  cross  bearing.  After  the  bolts  are  tightened  the  spike 
can  be  removed  and  holes  bored  for  the  pins  which  are  then 
driven  through. 

In  making  joints  in  woodwork  study  the  problem  carefully. 
Avoid  pockets  where  moisture  may  collect.  Cracks  seriously  weaken 
timber  framework,  and  as  cracks  usually  start  from  interior  angles 
all  complicated  framing  joints  should  be  avoided.  Frame  all  pieces 
so  the  center  lines  through  stressed  areas  will  meet  at  a  common 
point  and  try  to  make  all  jointing  lines  straight.  Jogs  in  joints 
not  only  are  apt  to  start  cracks,  but  there  is  a  danger  of  shrinkage, 
causing  unequal  bearing,  which  will  set  up  bending  moments 
about  the  joint,  so  it  is  best  to  have  straight  joints  which  admit  of 
adjustment.  If  through  any  mischance  the  two  faces  do  not  meet 
perfectly,  insert  a  thin  sheet  of  lead,  which,  in  course  of  time,  will 
equalize  the  bearing.  Engineering  principles  were  seldom  used 
for  timber  joints  in  former  generations,  much  of  the  work  being 
done  by  carpenters  who  had  no  engineering  training  and  by  drafts- 
men who  blindly  copied  old  examples.  Design  the  joint  in  accord- 
ance with  engineering  principles,  not  forgetting  that  wood  shrinks 
and  that  when  moist  it  rots.  The  joint  is  bound  to  be  all  right  if 
it  is  simple  and  the  computations  show  it  to  be  adequate  for  the 
work  it  must  do. 

In  studying  joints  of  members  they  are  considered  as  single 
sticks.  Methods  for  making  single  sticks  out  of  a  number  of  pieces 


JOINTS   AND   CONNECTIONS 


145 


will  be  shown,  but,  while  it  makes  very  little  difference  for  tension 
pieces,  it  is  bad  practice  to  use  several  pieces  to  form  compression 
members  when  single  sticks  can  be  obtained. 

In  Fig.  90  is  shown  the  truss  illustrated  in  Fig.  80,  the  computa- 
tions for  which  were  made  as  an  exercise.  We  will  now  proceed 
to  design  some  of  the  joints.  Assume  the  wood  to  be  Yellow 


Pine,  Grade  1  (Underwriter's  Code),  in  which  the  allowable  safe 

stresses  are  as  follows: 

Tension 1600  Ibs.  per  sq.  in. 

Compression  (end  bearing) 1200  Ibs.  per  sq.  in. 

Compression  (side  bearing) 350  Ibs.  per  sq.  in. 

Shear  (with  the  grain) 120  Ibs.  per  sq.  in. 

The  stresses  in  the  steel  are  as  follows: 

Tension 16,000  Ibs.  per  sq.  in. 

Shear 10,000  Ibs.  per  sq.  in. 

Bearing 20,000  Ibs.  per  sq.  in. 

For  the  strength  of  bolts  and  lag  screws  in  combined  shear, 
bearing  and  cross  bending,  see  pages  138  and  139. 

The  compression  in  L0  U\  is  35,950  Ibs.  and  the  tension  in 
IioLi  25,000  Ibs.  The  lower  chord  is  always  dimensioned  to  take 
care  of  the  maximum  tension,  which  in  this  case  is  45,000  Ibs. 
This  leaves  considerable  excess  material  in  the  end  panels,  which 
is  available  for  cutting  to  form  connections.  Similarly,  the  upper 
chord  is  dimensioned  for  the  maximum  compression  and  the  size  is 
uniform  throughout.  . 

Before  proceeding  to  design  the  connection  joints  at  the  ends 
of  a  truss  and  at  the  ends  of  panels  it  is  necessary  to  know  the  sizes 
of  the  members.  If  material  has  to  be  cut  out  for  the  formation  of 
joints  the  required  area  can  be  added  to  the  member  provided 
it  has  not  enough  excess  material  to  provide  the  necessary  area 
for  the  details.  Proceeding  in  this  manner  we  will  study  all  the 


146  PRACTICAL  STRUCTURAL  DESIGN 

usual  methods  for  making  joints  in  tension  and  compression 
members  and  then  design  the  joints. 

It  is  not  always  possible  to  get  a  single  stick  to  use  as  a  tension 
or  compression  chord  in  a  truss,  and  it  is  necessary  to  build  them 
up  in  nearly  every  instance.  If  a  single  stick  is  used  the  area  is 
increased  to  allow  for  the  hole  through  which  the  largest  vertical 
rod  passes.  In  the  truss  now  being  considered  the  sizes  of  the  rods 
are  found  as  follows: 

The  rod  at  Us  will  be  1  in.  diameter  if  the  threaded  ends  are 
upset,  and  if  the  ends  are  not  upset  the  diameter  will  be  If  ins. 
This  rod  may  be  smaller  as  the  stress  is  very  low,  but  it  is  not 
advisable  to  use  a  rod  of  smaller  diameter  for  such  a  length.  The 
stress  in  rod  at  UzLz  will  be  the  same  as  the  rod  at  Us,  for  the  stress 
is  15,000  Ibs.  and  the  allowable  stress  in  the  steel  is  16,000  Ibs.  per 

square  inch.  The  net  area  of  the  rod  at  U\L\  =     '        =  1.56  sq.  ins., 

lo,UUU 

and  we  will  use  a  rod  1|  ins.  diameter  if  the  threaded  ends  are  upset. 
If  upset  ends  are  not  used  the  loss  due  to  cutting  of  the  threads 
will  require  the  use  of  a  rod  If  ins.  diameter.  The  following  table 
gives  the  dimensions  of  rods  with  upset  screw  ends.  To  obtain 
the  diameter  at  the  root  of  the  thread  cut  in  a  bolt  use  the  figures 
in  the  second  column  for  the  outside  diameter  and  the  diameter 
at  the  root  of  the  thread  will  be  found  in  the  third  column.  The 
screw  threads  in  the  table  are  Franklin  Institute  standards.  In 
the  1913  edition  of  the  "  Carnegie  Pocket  Companion  "  the  tables 
for  upset  screw  threads  are  American  Bridge  Company  standards. 

REMARKS.  —  As  upsetting  reduces  the  strength  of  iron,  bars 
having  the  same  diameter  at  root  of  thread  as  that  of  the  bar 
invariably  break  in  the  screw  end,  when  tested  to  destruction, 
without  developing  the  full  strength  of  the  bar.  It  is  therefore 
necessary  to  make  up  for  this  loss  in  strength  by  an  excess  of 
metal  in  the  upset  screw  ends  over  that  in  the  bar. 

To  make  one  upset  end  for  5-inch  length  of  thread,  allow  6-inch 
length  of  rod  additional. 


JOINTS   AND   CONNECTIONS 


147 


UPSET  SCREW  ENDS    FOR   ROUND   AND   SQUARE    BARS 


Diameter  of  Round 
or  Side  of  Square 
Bar,  Inches 

ROUND  BARS 

SQUARE  BARS 

Diameter  of 
Upset  Screw 
End, 
Inches. 

Diameter  of 
Screw  at  Root 
of  Thread, 
Inches. 

1* 

IN 

£3 

Excess  of  Effec- 
tive Area  of 
Screw  End  over 
Bar.  Per  cent. 

Diameter  of 
Upset  Screw 
End, 
Inches. 

Diameter  of 
Screw  at  Root 
of  Thread, 
Inches. 

^ 

|l 

Excess  of  Effec- 
tive Area  of 
Screw  End  over 
Bar.  Percent. 

K 

% 

.620 

10 

54 

^ 

.620 

10 

21 

T9* 

X 

.620 

10 

21 

K 

.731 

9 

33 

6/8 

% 

.731 

9 

37 

1 

.837 

8 

41 

H 

1 

.837 

8 

48 

1 

.837 

8 

17 

X 

1 

.837 

8 

25 

IX 

.940 

7 

23 

H 

ix 

.940 

7 

34 

ilA 

1.065 

7 

35 

% 

IX 

1.065 

7 

48 

IX 

1.160 

6 

38 

H 

IX 

1.065 

7 

29 

IX 

1.160 

6 

20 

i 

IX 

1.160 

6 

35 

11A 

1.284 

6 

29 

lA 

IX 

1.160 

6 

19 

1% 

1.389 

51A 

34 

iy* 

VA 

1.284 

6 

30 

1% 

1.389 

51A 

20 

If  8 

m 

1.284 

6 

17 

IX 

1.490 

5 

24 

IX 

1% 

1.389 

5M 

23 

iys 

1.615 

5 

31 

IrV 

1H 

1.490 

5 

29 

iy* 

1.615 

5 

19 

IX 

ix 

1.490 

5 

18 

2 

1.712 

4K 

22 

If, 

i% 

1.615 

5 

26 

2ys 

1.837 

VA 

28 

m 

2 

1.712 

4^ 

30 

VA 

1.837 

VA 

18 

ih 

2 

1.712 

4^ 

20 

VA 

1.962 

VA 

24 

W* 

2^ 

1.837 

4^ 

28 

VA 

2.087 

VA 

30 

1H 

2^ 

1.837 

4^ 

18 

2*A 

2.087 

VA 

20 

iM 

2K 

1.962 

4^ 

26 

&A 

2.175 

4 

21 

m 

VA 

1.962 

4^ 

17 

VA 

2.300 

4 

26 

1% 

%% 

2.087 

4J* 

24 

VA 

2.300 

4 

18 

lit 

VA 

2.175 

&A 

26 

2M 

2.425 

4 

23 

2 

VA 

2.175 

4 

18 

2% 

2.550 

4 

28 

2  A 

25/8 

2.300 

4 

24 

2Ji 

2.550 

4 

20 

2^ 

2% 

2.300 

4 

17 

3 

2.629 

3H 

20 

2i\ 

m 

2.425 

4 

23 

3K 

2.754 

3K 

24 

2^ 

zys 

2.550 

VA 

28 

VA 

2.754 

3^ 

18 

2* 

VA 

2.550 

±1A 

22 

VA 

2.879 

3J^ 

22 

2^ 

3 

2.629 

3^ 

23 

&A 

3.004 

3H 

26 

3iV 

sy8 

2.754 

3^ 

28 

&A 

3.004 

3^ 

19 

2^ 

VA 

2.754 

3M 

21 

VA 

3.100 

3^ 

21 

2A 

3M 

2.879 

sy2 

26 

&A 

3.225 

3M 

24 

2^ 

VA 

2.879 

VA 

20 

&A 

3.225 

3Ji 

19 

2H 

&A 

3.004 

&A 

25 

m 

3.317 

3 

20 

2^ 

VA 

3.004 

&A 

19 

VA 

3.442 

3 

23 

2H 

&A 

3.100 

VA 

22 

&A 

3.442 

3 

18 

2% 

&A 

3.225 

VA 

26 

4 

3.567 

3 

21 

2if 

&A 

3.225 

&A 

21 

4^ 

3.692 

3 

24 

3 

&A 

3.317 

3 

22 

4^ 

3.692 

3 

19 

VA 

VA 

3.442 

3 

21 

4^ 

3.923 

2% 

24 

&A 

4 

3.567 

3 

20 

4^ 

4.028 

2^ 

21 

&A 

VA 

3.692 

3 

20 

4^ 

4.153 

2M 

19 

VA 

VA 

3.798 

2^ 

18 

148 


PRACTICAL  STRUCTURAL   DESIGN 


The  following  table  gives  weights  and  areas  for  square  and  round 
bars  and  rods. 

WEIGHTS  AND  AREAS  OF  SQUARE  AND  ROUND  BARS 
AND   CIRCUMFERENCES   OF   ROUND   BARS 

(One  cubic  foot  of  steel  weighing  489.6  Ibs.) 


Ill 

1 

Weight  of 
D  Bar 
One  Foot 
Long 

Weight  of 
OBar 
One  Foot 
Long 

Area  of 
D  Bar 
in  Sq.  Inches 

Area  of 
OBar 
n  Sq.  Inches 

Circumference 
of  O  Bar 
in  Inches 

0 

A 

.013 

.010 

.0039 

.0031 

.1963 

H 

.053 

.042 

.0156 

.0123 

.3927 

A 

.119 

.094 

.0352 

.0276 

.5890 

H 

.212 

.167 

.0625 

.0491 

.7854 

A 

.333 

.261 

.0977 

.0767 

.9817 

y* 

.478 

.375 

.1406 

.1104 

1.1781 

& 

.651 

.511 

.1914 

.1503 

1.3744 

YT. 

.850 

.667 

.2500 

.1963 

1.5708 

& 

1.076 

.845 

.3164 

.2485 

1.7671 

y* 

1.328 

1.043 

.3906 

.3068 

1.9635 

H 

1.608 

1.262 

.4727 

.3712 

2.1598 

*A 

1.913 

1.502 

.5625 

.4418 

2.3562 

H 

2.245 

1.763 

.6602 

.5185 

2.5525 

M 

2.603 

2.044 

.7656 

.6013 

2.7489 

ii 

2.989 

2.347 

.8789 

.6903 

2.9452 

l 

3.400 

2.670 

1.0000 

.7854 

3.1416 

TV 

3.838 

3.014 

1.1289 

.8866 

3.3379 

X 

4.303 

3.379 

1.2656 

.9940 

3.5343 

A 

4.795 

3.766 

1.4102 

1.1075 

3.7306 

X 

5.312 

4.173 

1.5625 

1.2272 

3.9270 

A 

5.857 

4.600 

1.7227 

1.3530 

4.1233 

N 

6.428 

5.049 

1.8906 

1.4849 

4.3197 

£ 

7.026 

5.518 

2.0664 

1.6230 

4.5160 

X 

7.650 

6.008 

2.2500 

1.7671 

4.7124 

A 

8.301 

6.520 

2.4414 

1.9175 

4.9087 

H 

8.978 

7.051 

2.6406 

2.0739 

5.1051 

H 

9.682 

7.604 

2.8477 

2.2365 

5.3014 

M 

10.41 

8.178 

3.0625 

2.4053 

5.4978 

H 

11.17 

8.773 

3.2852 

2.5802 

5.6941 

H 

11.95 

9.388 

3.5156 

2.7612 

5.8905 

« 

12.76 

10.02 

3.7539 

2.9483 

6.0868 

2 

13.60 

10.68 

4.0000 

3.1416 

6.2832 

A 

14.46 

11.36 

4.2539 

3.3410 

6.4795 

H 

15.35 

12.06 

4.5156 

3.5466 

6.6759 

A 

16.27 

12.78 

4.7852 

3.7583 

6.8722 

J4 

17.22 

13.52 

5.0625 

3.9761 

7.0686 

TV 

18.19 

14.28 

5.3477 

4.2000 

7.2649 

M 

19.18 

15.07 

5.6406 

4.4301 

7.4613 

A 

20.20 

15.86 

5.9414 

4.6664 

7.6576 

H 

21.25 

16.69 

6.2500 

4.9087 

7.8540 

JOINTS   AND   CONNECTIONS 


149 


SQUARE   AND   ROUND   BARS  —  Continued 


Thickness 
of  Diameter 
in  Inches 

Weight  of 
D   Bar 
One  Foot 
Long 

Weight  of 
O  Bar 
One  Foot 
Long 

Area  of 
D  Bar 
in  Sq.  Inches 

Area  of 
SBar 
.  Inches 

Circumference 
of  O  Bar 
in  Inches 

A 

22.33 

17.53 

6.5664 

5.1572 

8.0503 

H 

23.43 

18.40 

6.8906 

5.4119 

8.2467 

H 

24.56 

19.29 

7.2227 

5.6727 

8.4430 

X 

25.71 

20.20 

7.5625 

5.9396 

8.6394 

it 

26.90 

21.12 

7.9102 

6.2126 

8.8357 

7/g 

28.10 

22.07 

8.2656 

6.4918 

9.0321 

ti 

29.34 

23.04 

8.6289 

6.7771 

9.2284 

3 

30.60 

24.03 

9.0000 

7.0686 

9.4248 

A 

31.89 

25.04 

9.3789 

7.3662 

9.6211 

^8 

33.20 

26.08 

9.7656 

7.6699 

9.8175 

A 

34.55 

27.13 

10.160 

7.9798 

10.014 

M 

35.92 

28.20 

10.563 

8.2958 

10.210 

A 

37.31 

29.30 

10.973 

8.6179 

10.407 

W 

38.73 

30.42 

11.391 

8.9462 

10.603 

A 

40.18 

31.56 

11.816 

9.2806 

10.799 

/4 

41.65 

32.71 

12.250 

9.6211 

10.996 

A 

43.14 

33.90 

12.691 

9.9678 

11.192 

H 

44.68 

35.09 

13.141 

10.321 

11.388 

H 

46.24 

36.31 

13.598 

10.680 

11.585 

% 

47.82 

37.56 

14.063 

11.045 

11.781 

It 

49.42 

38.81 

14.535 

11.416 

11.977 

K 

51.05 

40.10 

15.016 

11.793 

12.174 

8 

52.71 

41.40 

15.504 

12.177 

12.370 

For  designing  plate  washers  the  following  table  will  be  useful. 
TANK  IRON   AND   STEEL,  WEIGHT   OF   SUPERFICIAL   FOOT 


Thickness 
in  Inches 

Weight  in  Lbs. 

Thickness 
in  Inches 

Weight  in  Lbs. 

Iron 

Steel 

Iron 

Steel 

A  =  .03125 

1.27 

1.30 

A  =    -3125 

12.63 

12.88 

A  =  .0625 

2.52 

2.57 

3A  =    .375 

15.16 

15.46 

A  =  -09375 

3.79 

3.87 

A  =    .4375 

17.68 

18.03 

Ys  =  .125 

5.05 

5.15 

1A  =    .5 

20.21 

20.61 

A  =  -15625 

6.32 

6.45 

T\  =    .5625 

22.73 

23.19 

A  =  -1875 

7.58 

7.73 

M  =    .625 

25.26 

25.77 

A  =  .21875 

8.84 

9.02 

H  =    -75 

30.31 

30.92 

H  =  -25 

10.10 

10.30 

H  =    -875 

35.37 

36.08 

A  -  .28123 

11.38 

11.61 

1      =1. 

40.42 

41.23 

The  low  temperature  (as  compared  with  iron)  at  which  steel  plates  have 
to  be  finished  causes  a  slight  springing  of  the  rolls,  leaving  the  plate  thicker 
in  the  center.  This,  combined  with  greater  density,  causes  steel  plates,  if 
kept  up  to  full  thickness  on  the  edges,  to  weigh  more  than  iron.  Both  iron 
and  steel  over  72  inches  wide  are  liable  to  run  even  heavier  than  the  weights 
given  above. 


150  PRACTICAL  STRUCTURAL  DESIGN 

Tables  of  sizes  and  weights  of  plates  and  bars  for  all  the  gauges 
in  use  are  given  in  all  the  steel  handbooks. 

The  areas  of  washers  will  be  as  follows,  the  cross  bearing  strength 

5000 

of  the  wood  being  350  Ibs.  per  square  inch;  U3  and  L3  =          =  14.3 

ooO 

sq.  ins.    The  washer  should  extend  across  the  chord,  which  we  will 
assume  is  8  ins.  wide,  so  the  width  will  be,  —7^-  =  1.78  ins.    Make 

o 

it  2  ins.  wide.    The  thickness  =  4  x  (2  x  35°)  =  140o  in.  Ibs.  bend- 

6 

ing  moment  considering  it  to  be  two  cantilever  beams,  one  on  either 
side  of  the  bolt  extending  to  the  edge  of  the  chord. 


The  thickness  (  =  =  =  0.5!  i,(make  it 


£  in.).  Tables  of  standard  square  and  round  washers  may  be  used, 
if  available,  but  the  area  must  be  sufficient  to  keep  the  stress  on 
the  side  of  the  wood  down  to  the  allowable  limit.  The  washer  may 
be  round  or  square,  but  it  is  best  usually  to  have  to  go  across  the 
width  of  the  chord.  If  square  the  above  washer  will  be  3.81  ins. 
X  3.81  ins.  If  round  the  diameter  will  be  4f  in. 
The  washers  at  Uz  and  Lz  will  have  the  following  area: 

'        =  42.9  sq.  ins.    If  extended  across  the  chord  the  dimensions 

will  be  5|  ins.  X  8  ins.  If  square  the  dimensions  will  be  6.6  ins.  X  6.6 
ins.  If  round,  the  diameter  will  be  1\  ins.  Assuming  a  washer  of 
steel  with  dimensions  o|  ins.  X  8  ins.  the  thickness  will  be  obtained 
by  using  the  following  formula: 

WLb 


in  which  t  =  thickness  in  inches, 

W  =  total  load  centrally  applied, 
L  =  length  of  plate  (width  of  chord)  —  bolt  hole, 
6  =  width  of  plate  =  bolt  hole, 
/  =  allowable  unit  fiber  stress. 

The  above  formula  is  good  also  for  cast  iron,  using  for  the  fiber 
stress  a  stress  which  is  an  average  of  the  allowable  tensile  and  com- 
pressive  stresses.  With  cast  iron  the  thickness  obtained  will  be 
at  the  edge  of  the  nut  on  the  end  of  the  bolt,  and  it  may  diminish 
to  one-half  this  thickness  at  the  edges.  For  very  large  washers  a 


JOINTS  AND   CONNECTIONS  151 

saving  can  be  made  by  using  ribs,  each  rib  being  considered  as  a 
cantilever  carrying  a  part  of  the  load,  shear  being  duly  taken  into 
account.  No  casting  should  be  less  than  f  in.  thick  and  sharp 
corners  should  be  avoided. 

The  washers  at  Ui  and  LI  =      '        =  71.5  sq.  in.    If  extended 
ooU 

across  the  chord  the  dimensions  will  be  8  in.  x  9  in.  and  this  is 
the  best  size  to  make  them,  for  if  square  or  round  they  will  pro- 
ject beyond  the  edges,  thereby  decreasing  the  bearing  area  and 
increasing  the  stress  on  the  wood.  The  thickness  will  be  com- 
puted by  the  formula  used  in  the  case  of  the  washers  at  joints 
C/2  and  Z/2- 

The  size  of  the  lower  chord  will  now  be  computed.    The  maxi- 
mum tensile  stress  is  45,000  Ibs.    The  allowable  fiber  stress  is 


1600  Ibs.     The  area  =  =  28.1  sq.  ins.     The  width  will  be 

lOUU 

assumed  at  8  ins.,  from  which  will  be  subtracted  2  ins.  on  account 
of  the  hole  for  the  largest  vertical  rod,  which  leaves  a  net  width  of 

28  1 

6  ins.     The  depth  =  —~-  =  4.7  ins.  "If  it  will  be  possible  to  use  a 

single  piece  of  timber  for  the  bottom  chord  we  can  use  a  5  ins.  X  8 
ins.  stick.  It  is  not  possible  that  a  single  stick  can  be  obtained  and 
we  will  take  it  for  granted  some  splicing  will  be  necessary,  which 
will  call  for  two  lines  of  f-in.  bolts  going  through  the  sides  of  the 
chord.  This  makes  the  thickness  4.7  +  (2  x  f  )  =  6.2  ins.  Using 
commercial  size  timbers,  this  will  make  the  chord  8  ins.  wide  and 

7  ins.  deep.    It  is  usually  best  to  have  the  depth  equal  or  exceed 
the  width  and  we  will  make  the  chord  8  ins.  X  8  ins. 

The  maximum  stress  in  the  upper  chord  is  40,000  Ibs.  and  the 

area  =      '        =  33.3  sq.  ins.    Assuming  a  width  of  8  ins.  and  sub- 

33  3        * 
tracting  2  ins.  for  the  hole  for  the  rod  the  depth  =  —^-  =  5.55  ins. 

A  shallow  beam  has  a  tendency  to  deflect  unduly  and  if  it  is  under 
compression  the  deflection  will  be  increased.  It  will  be  advisable, 
therefore,  to  increase  the  depth  and  as  this  will  decrease  the  breadth 
the  minimum  advisable  thickness  should  be  found. 

The  allowable  maximum  compressive  fiber  stress  is  based  on 
a  length  not  exceeding  15  times  the  least  thickness.  When  the 
length  is  greater  the  fiber  stress  is  decreased.  One-fifteenth  of 


152  PRACTICAL  STRUCTURAL  DESIGN 

10  ft.  =  8  ins.,  so  this  width  should  be  maintained.  The  depth 
should  not  be  less  than  this  amount,  which  fixes  the  size  of  the 
top  chord  at  8  ins.  X  8  ins.  and  as  the  piece  will  carry  nothing  but 
its  own  weight  the  deflection  will  not  create  anxiety,  for  by  thus 

40  000 

increasing  the  size  of  the  piece  the  fiber  stress  becomes  -^-±—  5-  =834 

0x8 

Ibs.  per  square  inch. 

In  the  truss  being  designed  the  loads  are  small  and  the  members 
are  small,  so  there  should  be  no  difficulty  in  getting  pieces  of  the 
dimensions  here  given.  There  will  be  then  no  additional  areas 
to  subtract  for  bolt  holes.  In  trusses,  however,  which  carry  heavy 
loads  in  which  several  pieces  must  be  used  to  form  the  members 
allowance  must  always  be  made  for  bolt  holes.  In  this  truss  it  is 
assumed  that  all  the  loads  are  concentrated  at  the  panel  points. 
If  rafters  rest  on  the  top  or  bottom  chords  they  must  act  as  beams 
to  carry  such  loads  and  must  be  designed  for  the  stress  thus  caused 
in  addition  to  the  direct  stress  caused  by  tension  in  the  lower 
chord,  if  the  rafters  rest  on  it,  or  to  the  stress  caused  by  com- 
pression in  the  upper  chord  if  the  rafters  rest  on  it. 

For  a  piece  acting  as  a  combined  tie  and  beam  or  acting  as  a 
combined  strut  and  a  beam  use  the  following  formula  to  obtain 
the  breadth  when  the  depth  is  assumed. 


in  which  6  =  the  breadth  of  the  piece, 

/  =  the  maximum  fiber  stress  (compression  for  the  upper 

chord,  tension  for  the  lower  chord), 
h  =  the  depth  of  the  piece, 
M  =  the  bending  moment  in  inch  pounds, 
D  =  the  total  direct  load  (compression  or  tension). 

In  practical  work,  in  calculating  a  rectangular  piece,  the  depth 
may  be  assumed  and  the  breadth  computed  to  take  care  of  M. 

Add  enough  breadth  to  carry  the  direct  load.  Or,  assume  a 
breadth  and  design  for  a  depth  sufficient  to  take  care  of  M,  and 
add  enough  breadth  to  take  care  of  the  direct  load. 

Fig.  91  shows  a  bolted  fish-plate  splice  used  in  connecting  sec- 
tions of  a  solid  piece  in  tension.  Formerly  this  was  done  on  the 
assumption  that  the  bolts  bent  and  they  were  designed  to  resist 
the  bending  moment.  The  moment  arm  was  equal  in  length 


JOINTS   AND   CONNECTIONS 


153 


to  one-fourth  the  thickness  of  the  middle  piece,  plus  one-half  the 
thickness  of  the  outer  piece,  or  splice  pad.  The  load  was  half  the 
total  tension  and  each  bolt  carried  its  proportionate  share.  This 
is  the  method  found  in  the  majority  of  text  books.  It  called  for 
very  large  bolts  and  made  a  heavy,  awkward-appearing  connec- 
tion. It  is  the  proper  method  to  use  for  pin  connections  (Fig.  86), 
the  size  of  the  pins  being  also  figured  for  direct  shear  and  the 
computations  for  end  lengths  and  distances  between  bolts  being 
computed  as  in  the  examples  following. 

The  method  for  fish-plate  joints  used  by  the  writer  and  shown 
in  Fig.  91  is  the  modified  fish-plate  joint  proposed  by  Mr.  Dewell, 


*W-l"Bolh,  WUMal/eab/eHtishersecKhBoffS. 

Side   Eleve 


Top  View 
Fig.  91. 

and  the  strength  of  the  bolts  is  based  on  the  table  on  page  139. 
Net  area  required  =      '        =  28.1  sq.  ins. 

JLtJUU 

Assume  the  splice  pads  to  be  2  ins.  thick  and  8  ins.  wide,  which 
gives  an  area  of  32  sq.  ins.  Two  1-in.  bolts  will  occupy  a  space 
2  ins.  wide,  with  a  bearing  of  4  sq.  ins.,  which  will  bring  the  splice 
pads  down  to  the  proper  net  size. 

From  the  table  on  page  139  the  strength  of  1-in.  bolts  in  double 
shear  with  2-in.  fish-plates  =  2460  Ibs. 
45,000 


Number  of  bolts  required  = 


2460 


18.3.    There  should  be  an 


even  number  of  bolts  and  we  should  not  use  a  higher  stress  than  has 
been  shown  to  be  safe,  so  this  points  to  the  use  of  20  bolts.  This 
means  20  bolts  on  each  side  of  the  joint,  for  the  entire  pull  must 
be  carried  into  the  fish-plates  and  then  be  carried  by  another  set 
of  bolts  back  from  the  fish-plate  into  the  main  piece  on  the  other 
side  of  the  joint.  The  total  number  of  bolts  therefore  will  be  40. 
There  is  a  transverse  tension  in  every  piece  which  tends  to  cause 


154 


PRACTICAL  STRUCTURAL  DESIGN 


the  wood  to  split  along  the  center  line  of  the  bolts.  The  amount 
of  this  tension  is  one-tenth  the  longitudinal  tension,  and  the  safe 
allowable  fiber  stress  is  one-tenth  the  safe  allowable  longitudinal 
stress.  The  resultant  is  a  shearing  stress  and  must  be  allowed  for 
in  spacing  the  bolts. 

The  shearing  stress  is  assumed  to  act  at  each  edge  of  each  bolt 
so  that  when  a  bolt  passes  through  a  2-in.  plank  it  exerts  a  shear- 
ing stress  on  four  inches,  as  shown  at  (c)  Fig.  90.  We  are  now  ready 
to  space  the  bolts  and  determine  the  length  of  the  splice  pads. 

The  total  shearing  area  in  direct  pull  =      ' 

07  K 

Spacing  of  bolts  for  shear  =  2Q      4      2   = 


375  sq.  ins. 

2.34  ins. 


.35 
1.00 


„       .  .     ,.  .          45,000x0.1 

Spacmg  required  for  transverse  tension  =  16Q  x  20  x  4  = 

Adding  diameter  of  bolts 

Required  spacing  of  bolts  3.69  ins. 

Bolts  will  be  spaced  3|  in.  staggered,  with  double  this  distance 
from  the  ends  of  the  splice  pads. 


r-    -  tv  Spaces  <§  •?  *t>-a                                         -     ?• 
_a.          ,s-          ,"-.          ,«,          _a.          rft.          ,&.          _&-          _a_       .  _&. 

ij          I          •!          i! 

! 

~<r> 

1    ^    1    °    1    *    1    ° 

45000  fo       ! 

II 

7^ 

[45000^ 

L      -J. 

*<*> 

TOT 


/6,  l£' Shear  Pins 


10,  V  Bo/fs 

20,  3/g'x 3.%"  Washers 


Side    Elevation. 
Fig.  92 


By  using  thicker  splice  pads  there  would  be  required  a  less 
number  of  bolts  and  the  spacing  would  have  been  closer.  It  Is 
not  pretended  that  the  details  here  worked  out  are  the  most 
economical,  for  only  the  methods  are  shown.  Each  designer  should 
work  out  several  details.  The  bill  of  material  and  labor  must  be 
made  out  for  each  and  unit  prices  applied  in  order  to  determine 
the  least  expensive  detail. 


JOINTS  AND   CONNECTIONS  155 

All  bolts  should  have  a  driving  fit.  This  may  be  obtained  by 
boring  all  holes  from  one  side  with  a  bit  the  right  size  to  assure 
a  driving  fit.  If  any  hole  is  large  then  use  the  next  size  larger 
bolt.  In  determining  the  working  loads  on  bolts  the  washers  played 
no  part,  so  on  bolts  used  for  this  type  of  fish-plate  joint,  standard 
washers  may  be  used,  drawn  tight  enough  to  insure  a  good  bear- 
ing. "  In  general  the  fewer  bolts  there  are  to  place,  the  less  will 
be  the  cost  for  labor,  and  the  more  certain  will  be  the  combined 
action.  Against  these  considerations  must  be  weighed  the  amount 
of  metal  in  the  bolts  and  the  availability  of  the  chosen  size.  Stock 
bolts  are,  of  course,  cheaper  than  special  sizes."  (Dewell.) 

In  Fig.  92  is  shown  a  shear-pin  joint  for  tension  members.  This 
is  very  reliable  for  thoroughly  seasoned  timber  but  should  not  be 
used  in  green  timber.  Any  shrinkage'  of  the  timber  will  allow  a 
slip  in  the  joint.  Hardwood  pins  may  be  used,  but  metal  is  better 
and  square  bars  or  round  are  equally  serviceable.  Iron  pipe  is 
generally  used.  The  pieces  are  cut  and  fitted  together,  then  spiked 
to  hold  them  in  position  during  the  fitting  of  the  bolts.  After  the 
bolts  are  driven  and  the  washers  drawn  tight,  holes  are  bored  and 
the  shear  pins  driven.  The  drawing  shows  the  shear  pins  vertical, 
but  they  may  just  as  well  be  horizontal,  if  the  designer  fears 
vertical  pins  may  fall  out. 

Assume  the  diameter  of  the  shear  pins  to  be  \\  ins.  and  the 
fish-plates  to  be  3  ins.  X  8  ins.  The  net  section  of  the  two  plates 
will  be  4.5  x  8  =  36  sq.  ins.  The  unit  tensile  stress  in  the  plates 

will  be  — ^ —  =  1250  Ibs.  per  square  inch,  which  leaves  the  plates 

safe,  as  the  allowable  stress  is  1500  Ibs.  per  square  inch. 
The  number  of  pins  required  is  fixed  by  the  bearing  area  on  the 

end  of  the  fibers  in  the  holes,  =  .  ,,  45o°0°1onn  =  6.25.   Use  8 

U./O  X  o  X  l^UU 

pins,  for  there  must  be  an  even  number. 

Total  shearing  area  required  =      '        =  375  sq.  ins. 

375 

Spacing  of  pins  for  shear  =  ^ 5  =  5.85  ins.    Adding  the  thick- 
et X  o 

ness  of  the  pins,  5.85  +  1.5  =  7.35  ins.     Make  it  8  ins.  center  to 
center. 

Subtract  from  the  thickness  of  the  fish-plate  one-half  the  thick- 
ness of  the  pins,  which  leaves  3  —  0.75  =  2.25  ins.  for  the  uncut 


156  PRACTICAL  STRUCTURAL  DESIGN 

portion,  the  tension  acting  through  and  considered  as  concentrated 
in  the  center,  which  is  thus  3  -  1.125  =  1.875  ins.  from  the  shearing 
joint  between  the  main  piece  and  the  fish-plates.  Add  to  this 
one-half  the  projection  of  the  shear-pin  into  the  main  member 
=  1.875  +  0.75  =  2.625  ins.,  which  is  the  moment  arm  for  the 
couple  acting  in  the  joint.  The  moment  =  2.625  x  22,500  =  59,063 
in.  lb.,  the  action  tending  to  raise  one  end  of  the  plate  from  its 
seat.  This  is  resisted  by  tension  in  the  bolts. 

Assuming  the  bolts  to  be  set  halfway  between  the  shear  pins 
the  length  of  the  moment  (or  lever)  arm  from  the  edge  of  the 
hole  to  the  center  of  the  bolt  =  4  -  0.75  =  3.25  ins.  The  stress 

59  063 
in  the  bolts  =      '        =  18,200  Ibs.     Four  bolts  will  be  used,  as 

oJaO 

18'  200 
shown,  and  the  stress  =  — ^ —  =  4550  Ibs.  per  bolt.     Use  f-in. 

bolts,  the  net  area  of  which,  at  the  root  of  the  threads,  =  0.42  sq.  in., 

4550 
which  causes  a  stress  =  -^5  =  10,830  Ibs.  per  square  inch,  which 

will  be  all  right  for  a  wrought  iron  bolt. 

For  developing  the  bolts  plate  washers  may  be  designed.  The 
tension  on  each  bolt  =  4550  Ibs.  and  the  area  for  each  washer 

=  — — -  =  13  sq.  ins.    Make  each  washer  3f  x  3f  ins.    The  student 
o5U 

can  compute  the  thickness  as  an  exercise  by  the  formula  on  page 
151 .  Standard  round  washers  of  equal  area  may  of  course  be  used. 
The  area  of  the  chord  will  now  be  checked.  Vertically  there 
will  be  a  hole  1|  diameter  (half  on  each  side),  which  subtracts 
12  sq.  ins.  Horizontally  there  will  be  two  f-in.  holes,  with  an 
area  of  11.375  sq.  ins.,  which,  added  to  the  area  of  the  vertical 
holes,  =  11.375  +  12  =  23.375  sq.  ins.  The  gross  area  of  the 
chord  is  64  sq.  ins.  and  the  net  area  =  40.625  sq.  ins.  The 

fiber  stress  =      '        =  1100  Ibs.  per  square  inch.    The  chord  has 

plenty  of  area  for  the  maximum  tension. 

In  Fig.  93  is  illustrated  an  old  type  known  as  a  tabled  fish- 
plate splice.  It  may  be  considered  to  be  reasonably  effective  when 
the  entire  stress  can  be  taken  by  not  more  than  two  tables  on 
either  side  of  the  chDrd  joint.  There  is  comparatively  little  sec- 
ondary tension  in  the  bolts,  therefore  they  can  act  in  their  most 
efficient  manner.  Washers  of  generous  size  must  be  provided  in 


JOINTS   AND   CONNECTIONS 


157 


order  that  the  joint  may  be  well  pulled  together  at  the  time  of 
framing  and  the  bolts  be  able  to  hold  the  tables  in  place  when  the 
stress  comes.  The  joint  is  dependent  to  a  very  large  degree  on 
the  tightness  with  which  the  timbers  are  held  in  place  by  the 
bolts,  and  excessive  shrinkage  in  the  timber  would  allow  the  fish- 
plates to  be  overstrained.  If  it  is  not  certain  that  the  timber  will 
be  well  seasoned  before  use,  the  fish-plates  should  be  made  larger 
than  computations  indicate  to  be  necessary  and  it  will  be  advisable 
to  use  about  ten  per  cent  more  bolts  than  those  provided  by  com- 
putations. Spikes  can  be  toe-nailed  into  the  fish-plates  and  will 
be  a  great  help. 

Depth  of  cut  for  table  and  chord:     Area  required  for  cut 
45,000 


1600  x  2  x  8 

Length  of  table  for  shear:  Area  required  = 


— 


=  22.4 


o  X  A  X 
ins,  (make  it  23  ins.) 

Size  of  bolts  required  :  The  stress  is  transmitted  from  the  uncut 
portion  of  the  chord  to  the  uncut  portion  of  the  fish-plate  past  the 


. 2,4"x8"5plice  Pads  ?8-4* 7- -> 

JL-.      f     [-$— u-l-4 


i 


t<- — 23"- < .-23"-— »<- 23— ><- 23 

Top    View. 


•— ±T  "^ f—  -f— - 

....Jfp.._.._J.._ ^-....S-//^ 


Side     Elevation. 
Fig.  93 

joint,  where  it  is  again  transmitted  to  the  other  section  of  the 
chord.  The  resultant  stress  thus  travels  through  the  center  of 
the  uncut  portion  of  the  fish-plate,  which  in  this  case  is  two  inches 
thick,  so  the  center  of  stress  is  1  in.  from  the  face.  The  resultant 
of  the  pressure  on  the  table  where  it  transfers  stress  is  at  the 
center  of  the  cut,  which  in  this  case  is  1  in.  from  the  inner  side  of 
the  fish-plate.  There  is  a  moment  arm  between  the  compressive 


158 


PRACTICAL  STRUCTURAL  DESIGN 


stress  on  the  edge  of  the  table  and  the  tensile  stress  in  the  uncut 
portion  of  the  fish-plate  which  is  equal  to  one-half  the  thickness  of 
the  fish-plate,  in  this  case  2  ins.  These  two  equal  and  opposite 
forces  constitute  a  couple  acting  on  the  fish-plate  equal  to  one-half 
the  stress  in  the  chord  times  one-half  the  thickness  of  the  fish- 
plate, or,  22,500  Ibs.  x  2  ins.  =  45,000  in.  Ibs.  This  moment 
must  be  resisted  by  tension  in  the  bolts  acting  about  the  bearing 
face  of  the  tables.  The  bolts  should  be  placed  on  the  vertical 
line  through  the  center  of  the  tables.  Their  lever  arm  is  thus 
equal  to  one-half  the  table  length,  in  this  case  11.5  in.  The  ten- 


sion on  the  bolts 


11. 


3920  Ibs.     Two  bolts  will  be  used 


having  an  area  of  0.3920  sq.  in.,  for  the  stress  on  wrought  iron  bolts 
should  not  exceed  10,000  Ibs.  per  sq.  in.  The  nearest  size  is  found 
to  be  |  in.  bolts,  which  have  an  area  at  the  root  of  the  threads 
=  0.202  sq.  in.  and  the  combined  area  of  two  f-in.  bolts  =  0.404 
sq.  in.  Two  f-in.  bolts  therefore  will  be  used  in  the  middle  of 
each  table  and  four  more  will  be  used  as  shown  to  bind  the  joint 
together. 

In  Fig.  94  is  shown  a  steel-tabled  fish-plate  joint.  This  is  a 
joint  that  requires  especially  good  and  careful  inspection.  It  is  a 
detail  for  members  carrying  heavy  stresses.  It  costs  considerable 
for  materials  and  on  account  of  the  number  of  -tables  required 
the  labor  cost  is  high,  for  there  must  be  very  careful  cutting  to 

insure     even     and 

tf$t-/af"-*$*-s~-9tr>p  9"^Jy/of~-^£fr  snuS  bearing  for  all 

the  tables. 

Bearing  area  re- 
quired for  tables  = 

.ins. 


,  . 

8,l%x8"Tables,bear/ngedgesmf//ed.AI/riVefs34  depth    of    tables  = 

12,  $sBolfs.  37  5 
Side    Elevation.  ~~ 


=  2>34    in> 


Fig'  94 


(make  it  2|  in.). 


Will  use  tables  IfV  in.  =  8  in.,  requiring  8  tables  in  all. 
Each  table  transmits  — ^ —  =  11,250  Ibs.  and  requires  three 
f-in.  rivets,  as  determined  by  bearing  on  a  TVin-  plate  (see  PaSe 


JOINTS   AND   CONNECTIONS  159 

190).  The  tables  are  solid  pieces  of  steel  bar  1T\  ins.  thick  by  3  ins. 
wide  riveted  to  the  i^-in.  plate  used  as  a  fish-plate.  The  thickness 
of  the  fish-plate  is  determined  as  follows: 

Net  section  of  one  plate  required  =  ~  -  '    ^^  =  1.41  sq.  ins. 

a    X 


Net  section  of  fy  by  8-in.  steel  plate  (deducting  the  three  rivet 
holes  for  the  table  rivets)  =  fV  X  (8  -  (3  x  f))  =  1.68  sq.  ins. 
Size  of  bolts  required  to  resist  moment  on  tables:    Moment 


=  11,250  Ibs.  x  |1  in.  =  7383  in.  Ibs.     Tension  in  bolts  =  - 

o.O 

=  2110  Ibs.     Use  two  f-in.  bolts. 
Space  between  tables  =  +  1|  =  13|  in. 

1  I  ,  —  *)() 

Sometimes  the  lower  chord  cannot  be  made  of  a  single  piece 
spliced  by  means  of  a  fish-plate  or  shear  pin  splices,  but  a  number 
of  2-in.  or  3-in.  planks  must  be  used.  The  methods  adopted  for 
splicing  such  chords  are  illustrated  in  Fig.  95. 

It  may  be  assumed  at  the  start  that  there  should  be  an  odd 
number  of  planks,  for  the  center  plank  has  so  much  section  cut 
away  for  holes  through  which  the  verticals  will  pass,  that  it  cannot 
be  counted  on  as  furnishing  any  tensile  strength.  Usually  the 
middle  rod  is  smaller  than  the  others  and  about  half  the  section 
of  the  center  plank  is  available  at  this  point  for  tensile  strength, 
which  is  an  additional  factor  of  safety.  The  center  planks  are 
regarded  merely  as  blocks  in  which  a  bearing  is  obtained  for  bolts. 

The  lower  chord  under  consideration  is  so  light  that  only  three 
planks  will  be  used.  The  center  plank  will  be  2  ins.  and  the  outer 
planks  will  each  be  3  ins.  thick.  It  is  well  to  have  the  planks 
rough  sawed  and  not  finished,  as  the  rough  surfaces  increase 
friction  and  also  leave  a  small  space  for  ventilation  between  the 
planks.  The  length  of  the  truss  is  60  ft.  center  to  center  of  bear- 
ings, and,  in  the  absence  of  data  as  to  the  end  joints,  we  will 
assume  two  feet  added,  making  the  total  length  62  ft.  The  outer 
planks  must  lap  past  each  other  for  some  considerable  distance 
to  allow  space  for  the  bolts  which  will  be  used  to  tie  them  to- 
gether. Assuming  that  planks  may  be  purchased  8  ins.  X  38  ft., 
each  side  will  have  one  plank  38  ft.  long  and  one  plank  62  —  38  =  24 
ft.  long.  These  lengths  will  be  alternated  so  that  at  one  end  on 
one  side  there  will  be  a  38-ft.  plank  and  on  the  opposite  side  of 
the  chord  there  will  be  a  24-ft.  plank.  This  gives  a  lap  of  14  ft. 


160  PRACTICAL  STRUCTURAL  DESIGN 

The  lap  will  be  in  the  panels  in  which  the  stress  is  45,000  Ibs. 
Each  outer  plank  will  carry  half  the  stress.  We  will  assume  two  rows 
of  f-in.  bolts  fastening  the  planks  together  and  these  bolts  will  be 
in  double  shear,  for  they  pass  through  the  center  plank.  Referring 
to  the  table  on  page  139  each  bolt  is  capable  of  carrying  1450  Ibs. 

While  each  outer  planks  carries  half  the  stress,  when  all  bolting 
is  done,  all  the  stress  passes  through  the  middle  splice,  half  coming 
in  from  each  side  plank  and  being  transferred  through  the  bolts 
to  the  plank  on  the  other  side;  therefore  the  bolts  in  the  middle 
splice  must  be  proportioned  to  carry  all  the  tension. 


Number  of  bolts  required  =  =  31.    Use  32  bolts. 


45,000 
Total  shearing  area  =  =  375  sq.  in. 

375 

Spacing  of  bolts  for  shear  =  •==  —  -  —  ^  =  0.975  in. 

oZ  X  O  X  ^ 

c.  •    it  •          45,000x0.1      „.,,„. 

Spacing  required  for  transverse  tension  =      '      00       =  0.147  in. 


Adding  diameter  of  bolts  =0.75 


Required  spacing  of  bolts  =  1.872  in. 

The  spacing  may  be  increased  to  any  desirable  amount,  so  before 
settling  the  matter  the  connections  for  the  other  planks  will  be 
taken  up. 

The  other  planks  each  carry  one-half  the  tension,  so  all  the  fig- 
ures above  may  be  prorated  accordingly.  In  each  joint  there  will 

375 
be  16  bolts.   The  total  area  required  for  shear  =  -^-  =  187.5  sq.  ins. 

z 

1 88 

Spacing  of  bolts  for  shear  =  -^ —  =  0.98    in. 

lo  x  t>  x  ^ 

22,500x0.1      A1._ 
Spacing  required  for  transverse  tension  =  — r~ — lfi       =  0.146 

Adding  diameter  of  bolts  =  0.75 


Required  spacing  for  bolts  =  1.872  in. 

Before  spacing  the  bolts  determine  how  far  the  first  bolts  will 
be  placed  from  the  end  of  the  planks.  The  shear  on  one  bolt  is 
1450  Ibs.  with  bearing  on  a  thickness  of  6  ins.  of  plank.  For  the 
interior  bolts  this  is  placed  on  two  lines  at  the  ends  of  the  diameter, 
but  for  an  end  bolt  it  is  best  to  use  but  one  line,  considering  that 
the  greatest  danger  is  shearing  on  a  line  through  the  center 
of  the  bolts.  The  length  required  for  shear  on  the  end  bolt 


JOINTS   AND   CONNECTIONS 


161 


1450 


2.02  in.,  so  the  end  bolts  may  be  placed  2.02  + 


0.75 
"2" 


6x120 
=  2.395  ins.  from  the  end  of  the  plank. 

In  Fig.  95  is  shown  (with  width  exaggerated)  the  arrangement 
of  the  bolts.  The  spacing  so  carefully  figured  is  the  closest  safe 
spacing.  The  bolts  can  be  placed  much  farther  apart  if  desired. 
In  the  middle  splice  carrying  all  the  tension  place  two  bolts  seven 
inches  from  the  ends  of  the  planks.  Space  the  remaining  bolts 
10  ins.  center  to  center.  The  14-ft.  lap  in  the  middle  of  the  truss 
is  then  connected  up  so  that  the  two  outer  planks  act  to  carry 


rnTTniiini 


Fig.  95 

half  the  tension  from  one  support  to  the  other.  It  now  remains 
to  bolt  the  other  two  outside  planks  to  them  in  order  that  the  rest 
of  the  tension  can  be  carried. 

There  should  be  two  bolts  close  to  each  joint,  so,  six  inches  from 
the  end  of  the  planks,  put  in  two  bolts.  Each  joint  has  16  bolts  in 
two  lines,  which  bolts  may  be  spaced  12  ins.  center  to  center.  For 
the  rest  of  the  chord  put  bolts  at  intervals  of  4  ft.  staggered  and 
at  the  very  end  put  two  bolts  through  6  ins.  from  the  end  to  make 
a  firm  bearing  for  the  end  of  the  brace.  Between  the  bolts, 
used  merely  to  hold  the  planks  firm,  large  spikes  may  be  driven 
at  intervals  of  about  12  ins.  to  make  the  whole  construction 
more  rigid. 

The  same  principles  apply  when  five  or  more  planks  are  used. 
In  such  a  case,  however,  it  often  happens  that  long  planks  may 
be  placed  in  the  middle  of  the  span  in  such  manner  that  all  the 


162  PRACTICAL  STRUCTURAL  DESIGN 

splices  will  come  in  panels  in  which  the  stress  is  low,  the  full 
area  of  the  planks  being  available  for  carrying  the  tension  hi  the 
middle  of  the  span  where  it  is  greatest.  In  any  event  the  splicing 
of  the  bottom  chord  of  a  truss,  when  said  chord  is  composed  of 
plank,  is  a  matter  requiring  a  great  deal  of  careful  study.  It  is 
an  easy  matter  to  make  a  poor  splice.  The  writer  has  seen  some 
in  which  the  designer  calculated  for  one-half  the  stress  going 
through  a  certain  splice  when  actually  all  the  stress  went  through  it. 
In  Fig.  96  is  illustrated  a  detail  of  a  bottom  chord  in  which  all 
the  tension  is  carried  by  a  steel  plate  and  underneath  are  two 
pieces  of  timber  large  enough  to  carry  the  plate  and  take  out  the 
sag.  This  is  from  a  plan  for  a  standard  wooden  highway  bridge 


.-/?,  Bo/fs 
fe 


IffOaA 


ir^^v:ir;ii?0/_:v/.:::rj:::::^/j/j::::z::;:;..-v^vZ?        Lower  Chord 
Bottom  View  of  Chord .  Cross  Section. 

Fig.  96 

designed  by  Hugh  C.  Lewis,  Bridge  Engineer  in  the  State  Highway 
Department  of  Utah,  E.  R.  Morton,  State  Road  Engineer.  The 
figures  were  copied  from  Engineering  News,  Sept.  21,  1916.  Note 
the  spliced  joint  between  the  two  timber  pieces  in  the  chord. 
This  consists  of  a  long  splice  plate,  large  enough  to  carry  all  the 
stress,  between  the  two  outer  chord  pieces.  The  use  of  such  a 
splice  leaves  a  wide  air  space  between  the  two  chord  pieces  for 
ventilation.  When  a  bottom  chord  consists  of  three  planks 
and  the  center  plank  between  two  vertical  rods  is  long  enough 
to  serve  as  a  splice  piece,  the  two  outside  pieces  may  break 
joints  at  a  section  passing  through  the  chord,  as  shown  in  this 
detail. 

In  Fig.  97  several  methods  are  shown  for  making  joints  in  a 
piece  under  compression.  These  joints  are  used  in  the  top  chords 
of  trusses  and  also  in  columns,  for  the  top  chord  of  a  truss  is  a 
column.  The  detail  shown  at  (a)  is  the  best  of  the  lot.  The  ends 
should  be  carefully  dressed  to  insure  an  even  bearing.  The  detail 
at  (6)  is  in  common  use  and  is  not  so  good  as  (a).  It  has  two  bear- 
ing surfaces  and  this  makes  it  very  difficult  to  get  an  absolutely 
true  bearing.  It  may  fit  tight  at  one  end  and  not  fit  evenly  at  the 
other  end.  When  the  load  is  brought  on  one-half  the  member 


JOINTS   AND   CONNECTIONS  163 

carries  all  the  load  until  the  fibers  give  and  then  the  other  end 
comes  into  bearing.  The  detail  of  (c)  should  never  be  used.  It 
has  two  end  bearings  and  in  addition  has  the  sloping  face  which 
is  difficult  to  fit.  If  one,  or  both  ends,  compress  through  rotting 
or  crushing  of  the  fibers,  the  load  is  carried  on  the  sloping  face, 
which  increases  the  tension  on  the  bolts  and  hastens  the  destruc- 
tion of  the  member.  When  splices  are  made  in  a  top  chord  the 
joints  are  preferably  vertical,  the  views  shown  being  top  or  bottom, 
as  may  be  desired.  The  number  of  bolts  to  use,  and  the  sizes, 
are  matters  determined  by  judgment  and  experience  in  the  three 
details  shown. 

The  detail  at  (d)  is  one  commonly  used  when  the  piece  under 
compression  is  made  of  several  pieces.  The  pieces  should  be  as 
thick  as  possible,  and  if  more  than  one  thickness  is  used  use  the 
thinner  pieces  inside  and  the  thicker  pieces  on  the  outside.  To 
design  such  a  member  consider  the  load  to  be  uniformly  distributed 
so  that  each  piece  carries  a  load  proportionate  to  the  area.  If  one 
piece  bends,  part  of  the  load  it  carries  must  be  transferred  to  the 
adjoining  piece  by  shear,  so  shear  pins  are  inserted  at  intervals 
of  15  tunes  the  thickness  of  the  thinner  of  the  pieces.  Divide  the 
total  load  by  the  number  of  planes  between  the  pieces,  that  is, 
by  the  number  of  pieces  less  one.  This  may  be  assumed  to  be 
shear  and  it  is  divided  by  the  number  of  shear  pins  in  one  joint  to 
determine  the  amount  of  bearing  for  each  pin.  From  this  the 
bearing  area  may  be  ascertained  in  the  manner  shown  for  the  shear 
pin  splice  and  the  table  fish-plate  splice  for  tension  members.  The 
bolts  are  close  to  the  pins  and  are  designed  to  take  tension,  the 
amount  of  which  is  ascer- 
tained from  the  moment 
caused  by  the  load  carried 
by  the  outer  pieces.  A 
compression  member,  or 
column,  carefully  designed 
according  to  the  above 
method,  should  be  about 
95  per  cent  as  efficient  as 

a  solid  piece  of  the  same  outside  dimensions.  The  ends  should 
be  carefully  dressed  to  insure  the  load  being  uniformly  carried  by 
all  the  pieces.  It  is  advisable  to  use  a  thin  sheet  of  lead  on  each 
end  of  the  member. 


164  PRACTICAL  STRUCTURAL  DESIGN 

For  many  years  it  has  been  accepted  as  true  that  if  a  piece 
under  compression  is  made  of  a  number  of  smaller  pieces  it  will 
not  act  as  a  solid  piece.  A  number  of  experiments  were  made  to 
determine  this  and  it  was  discovered  that  the  secret  lay  in  the 
connections.  Pieces  as  ordinarily  made  were  found  to  be  very 
deficient  in  strength.  Thin  planks  spiked  together  as  thoroughly 
as  the  ingenuity  of  the  experimenter  could  devise  proved  to  be 
almost  as  strong  in  small  specimens  as  solid  pieces.  It  is  not 
likely,  however,  that  in  actual  work  this  amount  of  nailing  will 
be  done.  Experiments  made  on  rather  large  columns  did  not  show 
up  so  well  as  experiments  on  smaller  columns. 

It  is  considered  to  be  not  the  best  practice  to  build  up  com- 
pression members  of  thin  pieces,  and  when  slender  pieces  are  used 
they  should  be  as  few  in  number  as  possible  and  shear  pins  should 
be  used  as  shown  in  Fig.  97  (d).  If  a  number  of  thin  planks  must 
be  used  they  should  be  spiked  together  by  gradually  building  up, 
no  expense  for  spikes  being  spared.  After  the  piece  is  built  up 
lay  wide  pieces  across  the  edges  and  spike  these  pieces  to  the  edge 
of  each  plank.  These  cross  pieces  will  cover  the  two  sides  from 
one  end  to  the  other  and  serve  to  call  each  plank  to  the  assistance 
of  all  the  others  in  case  there  is  any  bending. 

It  has  been  stated  that  the  compressive  fiber  stress  given  in 
specifications  for  wood  is  based  on  pieces  having  a  length  not 
greater  than  15  times  the  diameter  or  least  thickness.  When 
the  proportions  adopted  provide  for  a  more  slender  column  the 
following  formula  is  used  to  ascertain  the  reduced  fiber  stress  to 
be  used. 


in  which 

/"  =  reduced  unit  fiber  stress. 

/  =  unit  fiber  stress  for  pieces  having  a  length  not  exceeding  15d. 
L  =  length  of  post. 

d  =  diameter  of  round  post  or  least  dimension  of  rectangular  post. 

This  formula  is  used  in  Chicago  for  wooden  posts.    There  are 
several  formulas  in  common  use  for  finding  the  reduced  stress  to 
use  for  slender  posts  and  these  will  be  discussed  in  the  chapter 
dealing  with  columns. 
L   The  reason  it  is  difficult  to  get  several  pieces  to  work  properly 


JOINTS   AND  CONNECTIONS  165 

unless  thoroughly  connected  by  shear  pins  or  spikes,  is  that  the 
ratio  of  slenderness  reduces  the  load  carrying  capacity  in  a  greater 
degree  than  the  load  is  reduced  by  the  number  of  pieces.  For 
example  the  allowable  unit  fiber  stress  for  solid  piece  with  dimen- 
sions equal  to,  or  greater  than,  one-fifteenth  the  length  may  be 
1200  Ibs.  per  sq.  in.  Divide  the  solid  piece  into  four  slices  and  the 
ratio  of  slenderness  for  each  slice  (plank)  is  one-fourth  of  one- 
fifteenth  =  -J-Q  L.  By  the  above  formula  the  allowable  fiber  stress 
is  only  400  Ibs.  per  sq.  in.  Assume  that  the  whole  load  is  carried 
on  the  cross  section  at  1200  Ibs.  per  sq.  in.  provided  it  is  a  solid 
piece.  By  dividing  it  into  four  planks,  each  carrying  one-fourth 
the  load,  it  is  seen  the  total  carrying  capacity  of  the  four  slender 
pieces  is  only  one-fourth  the  total  load.  That  is,  the  four  pieces 
acting  separately  are  each  strong  enough  to  carry  one-sixteenth 
of  the  load.  By  nailing  them  together  at  intervals  they  are  made 
to  act  together  to  some  extent,  and  if  arrangements  are  made 
to  connect  them  together  rigidly  at  intervals  not  greater  than  one- 
fifteenth  the  thickness  of  each  piece  the  allowable  fiber  stress  is 
increased.  Putting  bolts  or  spikes  through  several  planks  is  not 
nearly  so  effective  as  spiking  them  together  by  internal  nailing, 
that  is  by  "  piling  "  them,  by  which  term  is  meant  nailing  each 
plank  to  a  lower  one,  the  spikes  passing  through  at  least  three 
after  three  are  assembled.  Two  should  be  spiked  together  and 
the  spikes  clinched.  Then  the  "  piling  "  is  done  by  adding  a  plank 
first  to  one  side  and  then  a  plank  to  the  other  side,  so  the  first 
two  planks  form  the  core.  To  hold  the  two  outside  planks  the 
cross  pieces  are  nailed  on  the  edges. 

Member  UzL^  has  a  length  of  14.14  ft.,  measured  from  center 
line  to  center  line  of  the  chords,  but  somewhat  less  in  the  clear. 
The  total  length,  however,  will  be  used  for  convenience  in  pro- 
portioning the  piece.  Assume  a  piece  4  in.  x  8  in.,  in  which  the 

14  X  12 

ratio  of  least  thickness  to  length  =  — - —  =  40.     This  is  much 

too  great.     Try  a  6  in.  x  6  in.  and  the  ratio  =  — J —  =  28, 

6 

which  is  still  large,  but  we  will  investigate  and  see  whether  the 
piece  will  do. 

First  find  the  fiber  stress.  This  is  usually  done  by  using  a 
straight-line  formula,  but  other  formulas  are  discussed  in  the 


166  PRACTICAL  STRUCTURAL  DESIGN 

chapter  dealing  with  column  design.     The  straight-line  formula 
used  for  wooden  columns  in  Chicago  is  as  follows: 

L 


in  which/  =  reduced  fiber  stress  per  sq.  in.  hi  compression. 

/  =  fiber  stress  used  for  columns  having  a  length  less 

than  15  tunes  the  diameter  or  least  thickness. 
L  =  length. 

d  =  diameter  or  least  thickness. 

When  L  is  in  feet,  d  is  in  feet;  and  when  L  is  in  inches,  d 
is  in  inches. 

Using  this  formula  for  the  case  under  consideration 

/'  =  1200  (l  -  g^g2)  =  1200  x  0.65  =  780  Ibs.  per  sq.  in. 

The  area  of  the  piece  is  36  sq.  ins.  and  the  total  working  strength 
=  36  X  780  =  28,000  Ibs.  The  actual  load  it  must  carry  is  only 
7190  Ibs.,  but  in  ordinances  and  specifications  the  maximum  ratio 
for  the  length  divided  by  least  width  (ratio  of  slenderness)  is 
30  and  the  6  in.  x  6  in.  piece  barely  comes  within  the  limit.  This 
ratio  is  for  vertical  posts,  whereas  sloping  posts  have  a  tendency 
to  bend  under  their  own  weight,  so  something  must  be  added  for 
additional  stiffness.  If  it  were  not  for  this  the  4  in.  x  6  in.  piece 
would  be  good,  as  it  has  a  safe  compressive  strength  of  18,240 
ins.  considered  as  a  vertical  post.  The  ratio  of  slenderness  of  40, 
however,  is  against  it. 

The  piece  U*L\  and  the  piece  C/iL0  will  be  made  8  in.  x  8  in. 
without  computation,  for  the  former  has  only  a  load  of  21,750  Ibs. 
to  carry  and  as  a  vertical  post  a  6  in.  x  6  in.  can  safely  carry 
28,000  Ibs.  The  additional  stiffness  secured  by  adding  two  inches 
to  the  breadth  and  thickness  saves  it  from  bending.  To  compute 
it  we  find  that  the  safe  fiber  stress  is  835  Ibs.  and  it  can  carry  as 
a  vertical  column  53,440  Ibs.;  therefore  this  size  will  do  for  the 
end  piece. 

When  joint  details  are  designed  it  may  be  discovered  that  some 
of  the  members  must  be  made  larger  to  allow  for  bolt  holes  good 
daps.  This  additional  area  may  be  added  at  the  tune,  but  the 
detailing  of  the  truss  should  proceed  in  the  order  here  followed,  so 
the  pieces  used  in  the  computations  may  be  reasonably  close  to 
the  actual  dimensions  finally  adopted. 


JOINTS   AND   CONNECTIONS 


167 


The 


Designing  Joints 

In  Fig.  98  is  shown  one  method  for  making  the  joint  L0. 
computations,  in  order,  are  as  follows: 

Depth  of  toe,   6  =  45°,  therefore  n  =  1200  X  0.45  =  540  Ibs. 
per  sq.  in. 

35  950 
Required  area  in  bearing  =      '        =  66.6  sq.  ins. 


/>/>    /> 

Required  depth  of  face  =  —  - 


8.33  ins. 


The  above  operations  involved  finding  the  fiber  stress  in  com- 
pression per  sq.  in.,  dividing  the  total  load  to  find  the  required 
area  and  then  dividing  the  area  by  the  width  of  the  chord  to  find 
the  required  depth  of  the  end  of  the  brace.  This  depth  being 
normal  to  the  angle  of  the  brace  we  divide  it  by  the  secant 
of  the  angle  and  find  that  the  vertical  depth  of  cut  in  the  chord 
=  8.33  -5-  1.4141  =  5.9  ins.  The  depth  of  the  cut  should  be  such 
that  below  the  point  there  will  be  enough  area  left  in  the  chord 
to  carry  the  tension.  Neglecting  the  middle  filler,  the  width  of 


the  chord  is  6  ins.,  so  the  depth  = 


25,000 


=  2.6  ins.    Practically, 


6x1600 

the  depth  of  the  cut  should  not  be  more  than  one-half  the  depth 
of  the  chord,  so  another  detail  should  be  selected. 

We  can,  at  this  point,  assuming  the  computed  depth  of  cut  is 
correct,  proceed  to  find  the  length  of  chord  projection  for  shear 
and  find  the  center  line 
of  support;  merely  to 
show  in  detail  the  nec- 
essary computations.  f<- 
For  this  purpose  the  I 
depth  of  the  vertical 
cut  in  the  chord  will 
be  assumed  to  be  four 
inches. 

Forces  are  assumed 
to  be  concentrated,  or 
to  act  along  the  center 
lines  of  stressed  mem- 
bers. In  the  end  piece  the  compression  is  all  acting  on  the  square 
end  face,  4  ins.  deep  x  8  ins.  wide  and  the  center  lines  of  the 


168  PRACTICAL  STRUCTURAL  DESIGN 

forces  are  as  shown  by  the  heavy  lines.  Below  the  end  piece  the 
uncut  area  of  the  chord  carries  all  the  tension,  and  this  acts  in 
the  center,  as  shown  by  the  heavy  arrow. 

The  two  forces  form  a  couple  with  a  moment  arm  of  4  ins. 
The  moment  =  4  x  25,000  =  100,000  in.  Ibs. 

If  the  center  line  of  the  support  is  placed  under  the  vertical 
line  dropped  from  the  center  of  the  end  face  the  bending  moment 
just  found  will  exert  a  tendency  to  open  the  joint,  because  of  the 
pull  around  the  lower  edge.  The  reaction  therefore  should  come 
under  the  intersection  of  the  center  line  drawn  through  the  face 
of  the  end  piece  and  the  line  through  the  center  of  area  of  ten- 
sion, as  illustrated  in  Fig.  98.  The  exact  position  may  be  computed 

by  dividing  the  moment  just  found,  by  the  reaction,       '        =  4  ins. 


which  is  the  distance  required  to  the  left.  It  is  merely  a  coincidence 
that  in  this  particular  example  the  reaction  equaled  the  tension 
in  the  end  panel  of  the  lower  chord. 

It  remains  to  find  the  chord  projection  for  shear,  the  length  a. 
The  width  of  the  chord  is  8  ins.  and  the  allowable  shearing  stress 
is  120  Ibs.  per  sq.  in.,  with  a  total  compression  load  of  25,000  Ibs. 


We  have  discovered  that  the  center  line  of  the  reaction  should 
be  4  his.  to  the  left  of  the  center  of  the  bearing  area  on  the  end 
brace.  This  is  5.42  ins.  to  the  left  of  the  lowest  point  of  the  end 
brace,  where  it  is  set  into  the  chord,  and  there  is  consequently  a 
projection,  26  —  5.42  =  20.58  ins.  beyond  the  center  of  the  sup- 
porting wall  or  column.  This  may  be  concealed  by  corbelling  out 
the,  brickwork  of  the  wall,  but  very  often  such  a  projection  is 
objectionable.  Designers  who  do  not  think  clearly  will  often  place 
the  support  under  the  very  end  of  the  projecting  piece,  and  this 
sets  up  bending  moments  which  will  cause  the  chord  to  break. 

Assume,  for  example,  in  the  present  instance  that  the  center 
line  of  the  support  is  6  ins.  from  the  end  of  the  projecting  chord. 
This  leaves  20  ins.  to  the  deepest  cut  and  18.58  ins.  to  the  vertical 
line  from  the  center  of  the  face  of  the  brace.  The  bending  moment 
=  18.58  X  25,000  =  464,500  in.  Ibs.  The  vertical  moment  arm 
between  the  compression  and  tension  area  is  4  ins.  and  the  area  of 
the  tension  side  below  the  cut  =  8  X  5.17  =  41.36  sq.  ins. 


JOINTS   AND   CONNECTIONS 


169 


464  500 

The  tensile  stress  = '          =  2807  Ibs.  per  sq.  in.,  and  the 

4  X  41. oo 

allowable  stress  is  1600  Ibs.  per  sq.  in.,  therefore  the  center  line 
of  the  support  cannot 
be  so  far  from  the 
end  of  the  brace.  The 
proper  position,  in  or- 
der to  keep  all  forces 
in  equilibrium,  is  5.42 
ins.  left  of  the  bottom 
point  of  the  end  piece. 
The  tension  caused  by 
moving  the  support 
farther  to  the  left 
must  be  added  to  the 
tension  in  the  chord; 
so  the  actual  stress,  if 
the  support  is  away  at  the  end,  =  (41.36  x  2807)  +  25,000  = 
141,098  Ibs. 

In  Fig.  99  is  shown  another  method  of  forming  the  joint,  the 
toe  cut  not  being  perpendicular  (normal)  to  the  line  of  thrust. 
The  angle  at  the  toe  is  75  degrees  and  the  angle  of  the  sloping 
bottom  of  the  cut  is  16  degrees;  for  the  angle  between  the  sur- 
faces may  vary,  it  not  being  necessary  to  have  the  lower  point 
a  right  angle.  Computations  will  be  made  to  obtain  the  area  of 
the  pressed  surface  and  the  allowable  and  actual  pressures  on  the 
surfaces.  The  computations  for  obtaining  the  projecting  length 
of  the  chord  for  shear  will  not  be  made,  for  it  is  like  the  work 
done  in  the  detail  shown  in  Fig.  98. 

Depth  of  toe:  6  =  75  degrees,  n  =  1200  x  0.75  =  900  Ibs.  per  sq.  in. 

35,950 

900 


Required  area  in  bearing 


40  sq.  ins. 


40 
Required  depth  of  toe  =  -3- 

o 


5  ins. 


This  depth  is  too  great  for  the  lower  chord,  for  if  it  is  used  the 
depth  of  the  chord  must  be  increased,  which  will  increase  the  weight 
and  be  an  uneconomical  proceeding.  We  will  therefore  abandon 
this  type  of  joint  for  the  truss,  but  the  computations  will  be  carried 
through,  merely  as  a  problem,  in  order  to  show  how  the  two  bear- 
ing surfaces  affect  the  position  of  the  center  line  of  the  support. 


170  PRACTICAL  STRUCTURAL  DESIGN 

Pressure  on  inclined  bed:  6  =  16  degrees,  n  =  1200  X  0.16  = 
192  Ibs.  per  sq.  in.  This  is  below  the  allowable  safe  pressure 
across  the  grain,  which  is  350  Ibs.  per  sq.  in.  which  we  will  use. 

18,000 
Required  area  in  bearing  =  =  51. o  sq.  ins. 

OOU 

Actual  area  =  10.5  x  8  =  84  sq.  ins.  The  area  therefore  is  more 
than  sufficient  for  its  component  of  pressure. 

The  student  is  to  observe  the  effect  the  distribution  of  the 
pressure  on  two  bearing  faces  has  on  the  depth  of  the  cut.  It  is 
obvious  that  when  the  cut  is  normal  to  the  stress  in  the  brace 
the  whole  thrust  must  be  taken  on  the  toe  of  the  post  and  none  is 
taken  by  the  inclined  face.  Actually  there  is  a  small  component 
normal  to  the  center  line  of  the  brace,  but  it  is  so  small  that  it  can 
be  neglected.  It  is  probably  taken  care  of  by  friction  of  the  toe 
on  the  cut,  or  by  slight  tension  in  the  bolts. 

The  small  diagram  in  the  upper  left-hand  corner  of  Fig.  99  is 
perhaps  self-evident,  but  will  be  explained.  The  load  travels 
down  the  brace  until  it  reaches  a  point  opposite  the  center  of  the 
inclined  face,  when  it  divides,  part  going  to  the  inclined  face  and 
part  to  the  toe.  This  is  drawn  to  scale  on  the  line  AB.  The  line 
AC  is  parallel  to  the  line  of  the  toe.  The  other  lines  require  no 
explanation,  for  the  values  are  marked  on  the  diagram  and  also 
on  the  drawing. 

The  vertical  component  of  the  inclined  face  is  18,000  Ibs.  and 
the  vertical  component  of  the  toe  is  10,500  Ibs.,  the  distance 

10  500 
between  them  being  4.5  in.    The  center  of  gravity  =  18000'+1Q50Q 

=  0.369  X  4.5  =  1.66  in.  from  the  right  component. 

The  bending  moment  of  the  couple  in  the  chord  =  4  x  25,000 
=  100,000  in.  Ibs.  This  divided  by  the  reaction  gives  the  distance 
the  center  line  of  the  support  must  be  shifted  to  the  left  to  insure 
equilibrium.  The  reaction  in  this  case  merely  happens  to  equal 
the  tension  in  the  lower  chord,  so  the  center  line  of  the  reaction 
must  be  4  ins.  to  the  left  of  the  position  above  found,  or,  4  +  1.66 
=  5.66  ins.  to  the  left  of  the  center  of  the  lower  bed  on  which  the 
brace  rests. 

Fig.  99  purposely  contains  as  few  lines  as  possible  in  order  to 
avoid  confusing  the  student.  In  both  details  there  are  two  inclined 
bolts  with  cast-iron  washers  to  keep  the  pieces  in  position,  so  all 
the  bearing  areas  will  have  a  proper  contact.  No  method  is  known 


JOINTS  AND   CONNECTIONS 


171 


whereby  the  stress  can  be  computed  in  these  bolts.  The  size  is 
fixed  by  judgment  based  on  experience.  For  a  truss  with  such 
light  loads  as  the  one  under  review,  f-in.  bolts  might  be  used, 
but  designers  generally  prefer  to  use  nothing  smaller  than  a 
f-in.  bolt,  this  being  good  practice.  Since  they  carry  a  very 
small  stress,  if  any,  it  is  not  necessary  to  cut  a  seat  in  the  lower 
chord,  which  would  weaken  it,  but  cast  iron  washers  of  the 
form  shown  are  used,  they  being  seated  in  the  timber  about  half 
an  inch. 

The  detail  in  Fig.  99  is  not  good,  for  the  reason  that  the  sloping 
face  on  the  16-degree  angle  cannot  be  accurately  cut  on  account 
of  the  depth  to  which  the  piece  must  go  into  the  chord.  The 
back  edge  of  the  sloping  face  should  meet  the  top  of  the  chord. 
These  details  must  be  worked  out  on  the  drawing  board  to  a  large 
scale  in  addition  to  being  computed.  The  computations  and  draw- 
ing to  scale  must  go  together.  In  all  the  end  joint  details  a  block 
is  inserted  through  which  the  bolts  go.  This  is  to  diminish  as  much 
as  possible  the  effect  of  secondary  stresses  set  up  by  the  bolts.  In 
Fig.  99  the  slope  of  the  bed  might  be  carried  to  the  top  of  the 
chord  without  weakening  it  and  a  block  fitted  in  the  space,  but  it 
increases  the  labor  cost.  The  designer  understands,  of  course, 
that  he  should  make  several  designs  and  choose  the  one  that  costs 
the  least  and  will  do  the  work. 

In  Fig.  100  is  illustrated  a  very  common  form  of  joint  and  it  is 
not  a  good  one.  This  j oint 
is  used  in  an  attempt  to 
get  rid  of  the  long  end  pro- 
jection caused  by  design- 
ing to  resist  shear.  The 
end  of  the  brace  is  dapped 
into  the  top  of  the  chord 
merely  to  hold  it  in  place. 
The  diagonal  bolts  take 
all  the  stress  and  the  shear 
is  resisted  close  to  the  bot- 
tom of  the  chord.  Note 
the  line  diagram  in  the 
upper  left-hand  corner  of 


Fig.  100 


the  figure.    The  line  A  B  is  parallel  with  the  brace  and  is  drawn 
to  a  scale  to  represent  the  load.     The  line  AC  is  parallel  to  the 


172 


PRACTICAL  STRUCTURAL  DESIGN 


direction  of  the  bolts.  The  line  BC  is  perpendicular  to  the  bed 
cut  into  the  top  of  the  chord.  The  line  BC  intersects  the  line 
AC  at  C.  The  line  AC  to  scale  gives  the  tension  in  the  bolts. 
This  is  found  to  be  45,500  Ibs. 

Deciding  to  use  two  bolts  the  tension  in  each  =  22,750  Ibs., 
which  calls  for  two  If-in.  bolts.  The  holes  must  be  filled  so  there 
is  no  reason  to  use  bolts  with  upset  threads,  and  the  size  of  the 
bolt  is  determined  by  the  area  at  the  root  of  the  thread.  The 


shear 


25,000 


13  ins.,  therefore  the  clear  distance  between 


2  x8  X  120 

the  edges  of  the  cuts  to  receive  the  bolts  cannot  be  less  than  13  ins. 
The  distance  from  the  edge  of  the  first  cut  to  the  end  of  the  chord 
must  be  equal  to  or  exceed  13  ins.,  for  each  bolt  washer  is  assumed 
to  carry  half  the  total  shear. 

The  allowable  pressure  on  the  side  of  the  wood  is  350  Ibs.  per 
sq.  in.  and  the  washers  on  the  face  of  the  end  brace  are  dimen- 

22  750 

sioned  as  follows:  ^ — '         =  8.1  ins.     Make  them  8  in.  x  8  in. 
o  X  ooO 

The  thickness  may  be  found  by  the  rules  mentioned  earlier.  The 
washers  on  the  lower  end  can  be  smaller,  for  they  bear  on  the  wood 
at  an  angle.  The  allowable  pressure,  the  angle  being  45  degrees, 
=  1200  X  p.45  =  540  Ibs.  per  sq.  in.  The  width  of  the  washers 

22  750 
=  — L-r-?7r  =  5.25  ins.   The  vertical  cut  into  the  bottom  of  the  chord 

o  X  o4(J 

to  permit  these  washers  to  bear  properly  is  3f  ins.  deep,  which  leaves 

a  space  of  3  ins.  be- 
tween the  top  of  the 
cut  and  the  bottom 
of  the  brace  seat  for 
the  tension  in  the 
chord.  Counting  the 
width  as  being  6  ins. 
the  strength  of  the 
chord  =  3  X  6  X  1600 
=  28,800  Ibs.,  so  the 
CLofffescf,™  chord  is  safe.  When 

Flg-  101  this  form  of  joint  is 

detailed  so  part  of  the  load  is  carried  by  thrust  against  the  chord 
and  part  is  carried  by  the  bolts,  the  weaker  detail  will  give  way 
before  the  other  is  brought  into  play,  as  they  do  not  act  together. 
Therefore  an  end  joint  should  be  designed  so  all  the  load  is  carried 


'2,l"BoIh 


JOINTS   AND   CONNECTIONS  173 

by  the  thrust  of  the  brace  against  the  chord,  or  it  must  all  be 
carried  by  the  bolts. 

In  Fig.  101  is  illustrated  a  low-cost  end  joint.  A  bolster  is 
placed  on  top  of  the  lower  chord  and  pins  are  used  to  transfer 
the  stress  by  shear,  2-in.  pins  being  used.  There  may  be  unequal 
shrinkage  in  the  chord  and  bolster  which  will  interfere  with  the 
proper  action  of  the  round  pins,  so  it  is  better  to  omit  the  pins 
and  use  a  bolster  large  enough  to  permit  it  to  lock  into  the  top 
of  the  chord  as  a  tabled  fish-plate.  The  vertical  bolts  carry  one- 
half  the  tension  and  the  method  to  use  in  figuring  the  size  of  the 
bolts  has  been  presented. 

This  joint  is  designed  as  follows:  Using  an  8  in.  x  10  in.  bolster 
the  depth  below  the  bottom  of  the  cut  will  be  4.34  ins.  Deduct 
8  sq.  ins.  for  the  half  pins  and  f  x  4.34  ins.  for  the  bolt  holes,  leaving 

25  000 
an  area  of  23.47  sq.  ins.    The  tensile  stress  will  be     '        =  1065  Ibs. 

per  sq.  in.,  and  the  allowable  safe  fiber  stress  is  1600  Ibs.  per  sq.  in. 
Therefore  the  8  in.  x  10  in.  bolster  is  O.K. 

OK  000 
Uncut  projection  for  shear  =  ~  —  -  -^  =  26  ins. 

o  X  l^JU  , 

The  compression  on  the  round  pins  will  be  taken  at  800  Ibs. 

25  000 
per  sq.  in.    The  required  number  of  2-in.  pins  =       '         =  3.9. 

Use  4  pins. 

The  thickness  of  bolster  back  of  the  brace  required  for  ten- 

25  000 

sion  =  ^'  '          =  2  ins.    The  thickness  is  4.34  ins.  less  1  in.  for 
o  X 


the  pins  =  3.34  ins.,  so  this  is  O.K. 

OK  000 
The  stress  per  pin  =  ^^-  =  6250  Ibs. 

The  clear  space  between  pins  =  5  -  —^  =  6.5  ins. 

o  X  l^&U 

Sometimes  a  detail  similar  to  that  shown  in  Fig.  100  is  used, 
but  instead  of  two  bolts  one  is  used.  In  the  case  considered  this 
bolt  will  be  2|-in.  diameter.  Instead  of  making  a  triangular  cut 
in  the  bottom  of  the  chord  to  form  a  bearing  surface  for  the  washer, 
a  casting  is  used  at  the  bottom  for  the  lower  end  of  the  bolt.  It 
is  very  common  to  use  such  washers  without  computing  the  size 
necessary,  and  in  many  existing  trusses  this  detail  is  weak.  The 
computations  are  as  follows,  referring  to  Fig.  102  (a)  : 


174 


PRACTICAL  STRUCTURAL  DESIGN 


In  the  diagram  in  the  upper  left-hand  corner  the  tensile  stress 
in  the  bolt  is  45,500  Ibs.  This  at  the  bottom  of  the  chord  is  con- 
verted into  a  horizontal  and  a  vertical 
component,  shown  in  the  lines  AC  and 
CD.  First  taking  the  value  given  by 
the  line  AD  we  find  the  depth  of  cut 
necessary  for  the  vertical  component. 


Depth  of  cut  = 


Fig.  102. 
=  3.34  ins.  (make  it  3|  ins.). 


The  bottom  of  the  plate  must  be  large  enough  to  allow  a  bearing 
of  350  Ibs.  per  sq.  in.  against  the  side  of  the  wood,  the  width  of 
the  plate  being  8  ins. 


Length  of  plate 


=  11.  4  ins.  (make  it  11%  ins.). 


X  oOU 

The  thickness  of  this  plate  must  be  found  by  assuming  the 
edges  projecting  beyond  the  collar  around  the  bolt,  as  cantilevers 
uniformly  loaded.  The  thickness  of  the  sloping  leg  in  the  chord 
must  be  figured  in  the  same  way.  The  plate  on  top  of  the  bracer 
at  the  upper  end  of  the  bolt  must  have  enough  area  to  keep  the 
bearing  down  to  350  Ibs.  per  sq.  in. 

The  plate  must  be  set  far  enough  from  the  end  of  the  chord  to 

prevent  shearing. 

oo  000 

Length  for  shear  =  ^    ^=  33.4  ins.  (make  it  33|  ins.). 

o  X  l^U 

The  center  of  the  bolt  should  be  as  nearly  as  possible  at  the 
center  of  the  plate,  which  helps  to  fix  the  position  of  the  bolt 
through  the  chord  and  brace,  taking  into  consideration  the  size  of 
washer  at  the  upper  end.  Insert  a  hardwood  block  in  the  angle. 

In  Fig.  102  (6)  is  a  detail  used  to  prevent  the  cutting  of  the 
bottom  chord  for  washers  and  to  avoid  the  expense  of  the  special 


JOINTS   AND   CONNECTIONS 


175 


casting  shown  at  (a).  The  position  of  the  bolt  is  fixed  by  the  size 
of  the  washer  at  the  upper  end  and  by  the  necessity  for  having 
the  uncut  portion  of  the  block  long  enough  to  resist  failure  by 
shear.  The  pins  are  shown  square,  but  they  may  be  round  if 
desired.  From  the  examples  given  the  student  should  have  no 
difficulty  in  designing  a  detail  such  as  this.  Supply  enough  pins 
for  bearing  and  enough  space  between  them  for  shear.  The  ver- 
tical bolts  are  in  tension,  but  of  course  this  tension  is  greatly  re- 
duced by  the  stiffness  of  the  block,  the  thinner  part  of  which 
must  have  area  enough  to  carry  the  tension  in  the  end  panel  of 
the  chord.  The  lower  washer  must  have  area  enough  to  keep 
the  pressure  to  the  limit  imposed  by  the  angle  at  which  the  pres- 
sure is  delivered  to  the  wood. 

In  Fig.  103  is  illustrated  a  cast-iron  shoe.  It  may  have  slightly 
different  details,  this  being  true  of  every  design  here  illustrated. 
The  form  shown  is  a  rather  com- 
mon type.  There  are  no  diagonal 
bolts,  so  the  pressure  on  the  top  of 
the  chord  is  not  uniform.  A  toe 
projecting  in  front  of  the  brace  is 
provided  to  take  care  of  this.  The  E^ 
depth  of  the  lugs  is  fixed  by  the  end 
bearing  strength  of  the  wood.  The 
thickness  of  the  lugs  is  fixed  by 
their  resistance  to  shear  and  bend- 
ing, for  they  are  short  cantilevers 
and  are  so  designed.  The  spacing 
of  the  lugs  is  governed  by  the 
shearing  strength  of  the  timber  Fig.  103. 

with  the  grain.  The  first  lug  is  usually  placed  directly  under  the 
end  of  the  batter  post.  The  arrangement  of  the  ribs  fitting  into 
the  end  of  the  batter  post  is  much  a  matter  of  judgment.  Cast 
iron  is  a  brittle  metal  and  it  is  best  to  have  the  least  thickness 
more  than  half  an  inch.  The  thickness  to  use  will  usually  be 
determined  by  the  size  of  the  casting;  a  large  heavy  casting  re- 
quiring larger  individual  parts,  should  have  all  the  parts  thick, 
because  a  fall  will  be  more  apt  to  injure  it  than  a  smaller  casting. 

Proceeding  with  the  design.  The  wood  at  the  end  of  the  batter 
post  is  not  held  by  bolts,  nor  confined  in  a  shoe,  therefore  a 
reduced  stress  will  be  used. 


176  PRACTICAL  STRUCTURAL  DESIGN 

Depth  of  toe,  0  =  45°.    n  =  540  Ibs.  per  sq.  in. 

_.       .     ,  ..          ,  ,       .          25,000 

Required  area  for  end  bearing  =  -^777-  =  46.4  sq.  ins. 

o4U 

Required  depth  of  vertical  cut  =  —  ~-  =  5.8  ins. 

Required  area  of  horizontal  cut,  6  =  45°.   n  =  540  Ibs.  per  sq.  in. 
Use  vertical  component  of  the  diagonal  load,  which  for  45  de- 
grees is  one-half. 

35,950       00  , 
Area  =  2^540  =  33'3  Sq"  mS' 

oo  o 

Length  of  horizontal  cut  =  5  -  -7^  —  -^  =  5.5  ins. 
o  —  (6  X  •§) 

In  this  example  the  angle  is  45  degrees  and  the  vertical  cut 
and  horizontal  cut  will  be  equal,  each  being  8  x  0.707  =  5.66  ins. 
The  vertical  cut,  for  the  assumed  fiber  stress,  should  be  5.8  ins., 
which  indicates  that  the  casting  should  be  designed  with  the 
two  bearing  surfaces  forming  an  angle  differing  enough  from 
90  degrees  to  keep  the  stress  within  the  limits  fixed.  If  the  design 

is  not  altered  the  stress  on  the  end  area  =  -=-^  —  -  =  552  Ibs.  per 

O.OD  X  o 

sq.  in.,  which  is  an  increase  of  only  4  per  cent,  so  will  be  allowed 
to  stand. 

The  maximum  unit  bearing  pressure  of  the  shoe  on  the  lower 
chord  must  not  exceed  the  allowable  bearing  stress  on  the  side 
of  the  fibers.  Draw  a  horizontal  line  from  the  mid-height  of  the 
toe  to  intersect  with  the  diagonal  line  meeting  the  point  of  the 
toe.  From  this  intersection  drop  a  vertical  line  to  represent 
the  vertical  component  of  the  thrust.  The  diagonal  line  rep- 
resents the  diagonal  thrust  and  the  horizontal  and  vertical 
components,  respectively,  are  shown  as  heavy  arrows. 

The  distance  from  the  vertical  component  to  the  front  edge 
of  shoe  (point  of  maximum  pressure)  scales  8.5  ins.  and  this  length 
will  be  called  a.  Next  find  the  length  of  the  shoe.  The  lugs  each 
carry  one-half  the  shear,  or  12,500  Ibs. 

12  500 
The  area  for  bearing  =      '     •  =  10.4  sq.  ins.    The  depth  of  the 


10  4 

lug  =  —^—  =  1.3  ins.  (make  it  1.5  ins.). 

o 

12  500 
Length  required  for  shear  =       '    ort  =  13  ins. 


JOINTS   AND   CONNECTIONS  177 

This  fixes  the  clear  distance  between  lugs  at  13  ins.  and  the 
front  lug  will  be  set  not  less  than  13  ins.  from  the  end  of  the  lower 
chord.   The  end  of  the  shoe  will  extend  2  ins.  beyond  the  rear  lug. 
This  makes  the  total  length  of  the  shoe,  L  =  25  ins. 
Let  a  =  distance  from  front  lug  to  toe  of  shoe. 
L  =  length  of  shoe. 

q  =  maximum  pressure  in  Ib.  per  sq.  in.  at  front  edge  of  toe. 
vertical  reaction      35,950  _  „ 

" 


which  is  well  within  allowable  stress. 

The  thickness  of  the  lugs  is  determined  by  treating  them  as 
cantilevers.    The  depth  is  1.5  ins.  loaded  uniformly  with  12,500  Ibs. 


The  bending  moment  =  f         %         )  X  12,500  =  14,080  in.  Ibs. 

The  moment  of  resistance  of  a  rectangular  section,  M r  =  Rbd?. 

The  compressive  stress  of  cast  iron  is  10,000  Ibs.  per  sq.  in.  and 
the  tensile  stress  is  3000  Ibs.  per  sq.  in.  A  mean  of  these  values 
may  be  used  in  determining  the  resisting  moment,  but  it  is  better 
to  use  the  tensile  stress,  and  R  =  3000  -=-  6  =  500.  Then 

=  V  500  x  8  ~ 
Moment  of  rotation  of  lugs  =  bending  moment  on  lug  =  14,080 

in.  Ibs.    Tension  in  bolt  back  of  lug  =  (2  '  .,„.  =  5120  Ibs.     Use 

one  |-in.  bolt. 

Figures  104,  105,  and  106  are  reproduced  from  an  article  by 
Henry  D.  Dewell  in 
Western  Engineering  for 
September,  1916,  the 
fourth  in  the  valuable 
series  referred  to. 

The  computations  for 
Fig.  104  are  as  follows: 

Area  required  for  bear- 
ing between   upper   and 


Fig.  104. 
sq.  ins.     (Mr.  Dewell  used  side  bearing  stress  of  285  Ibs.  per  sq.  in.) 


178  PRACTICAL  STRUCTURAL  DESIGN 

Depth  of  lugs  =  2  x  1600  x  8  =  1>915  inS'    Use  2"in' lugs' 
Thickness  of  lugs  for  bending : 

Bending  moment  on  one  lug  =  24,500  Ibs.  x  1.5  =  36,750  in.  Ibs. 
Thickness  of  lug  assumed  to  be  1  in. 

Required  section  modulus  =  OK'QQQ  =  1-465. 

Required  thickness  of  lug  =  1.05  ins.    Use  1-in.  plate  as  assumed. 

49  000 

Length  required  for  shear  between  lugs  =  = ~ — 5  =  20.4  ins. 

A  X  loU  X  o 

Use  1  ft.  8.5  ins. 

4-Q  000 

Depth  of  toe  =    *%,      Q  =  3.84  ins.     Use  4  ins. 
loUU  X  o 

Bearing  stress  of  1600  Ibs.  per  sq.  in.  is  used,  as  the  timber 
fibers  are  confined  and  therefore  capable  of  taking  full  end  com- 
pression. 

Stress  in  bolster  =  horizontal  component  of  stress  in  two  f-in. 
bolts  =  4830  Ibs.  x  2  x  0.5  =  4830  Ibs. 

4830 
No.  of  shear  pins  required  =  Qftn      0  =  0.75.   Use  one  2-in.  pin. 

oUU  X  o 

In  the  cast-iron  shoe  shown  in  Fig.  103  the  projection  of  the 
toe  limited  the  bearing  stress  on  the  top  of  the  chord.  In  the 
form  shown  in  Fig.  104,  since  the  line  of  thrust  intersects  the  shoe 
practically  at  the  end  of  the  toe,  the  inclined  bolts  will  be  called 
into  play  to  prevent  a  rotation  of  the  shoe,  which  would  greatly 

increase  the  toe  pres- 
sure. The  bolts,  there- 
fore, must  be  always 
tight  in  order  to 
secure  an  approxi- 
mately uniform  ver- 
pressure. 


|R         eWBotsfer    8,%'BoHs  This  form  of  joint 

Z^%*3^Y3ifJfcrfw*  10,  %6'x3%j(3%  Washers    js     considered     very 

Mill  Bearing  Edge  of  Tables.  . 

good    because    it    is 
simple  in  action  and 

comparatively  easy  to  frame  into  the  timber.  Both  lugs  must 
have  an  even  bearing  against  the  wood  and  this  is  a  hard  thing 
to  secure,  but  if  the  inspection  is  good  the  work  will  be  all  right. 
When  there  is  more  than  one  bearing  surface  this  objection 


JOINTS   AND   CONNECTIONS  179 

always  arises.  If  the  fitting  is  not  properly  done  the  joint  can 
be  shimmed  with  thin  metal  shims. 

Mr.  Dewell  makes  the  following  comments:  "This  shoe  can 
be  used  only  for  stresses  requiring  not  more  than  two  lugs,  hence 
its  field  of  application  is  limited.  Another  defect  is  that  the 
forge  work  is  difficult  with  the  thickness  of  plate  used.  Especially 
is  this  true  of  the  bending  of  the  end  of  the  inner  plate  to  form 
the  inner  lug.  Incidentally,  this  detail  forms  a  good  example  of 
the  consideration  of  actual  unit  working  stresses  as  compared 
with  purely  theoretical  values,  as  mentioned  in  the  first  article 
of  this  series.  With  a  2-in.  depth  of  lug,  the  bearing  pres- 
sure against  the  ends  of  the  fibers  is  assumed  to  be  1600  Ibs. 
per  sq.  in.  On  account  of  the  fillet  formed  in  bending 
the  plate,  the  actual  bearing  area  will  be  decreased  and  the 
actual  unit  working  stress  will  probably  be  around  1800  Ibs.  per 
sq.  in." 

In  Fig.  104  and  Fig.  105  the  sizes  of  the  two  diagonal  bolts  are 
determined  by  judgment  and  experience.  They  are  not  suscep- 
tible of  computation.  The  vertical  bolts  are  found  by  computa- 
tion. When  any  computation  is  omitted  in  any  of  the  examples 
it  is  for  the  reason  that  the  student  is  assumed  to  know  how  to 
make  it.  Every  detail  must  be  investigated  according  to  the 
principles  and  methods  illustrated. 

Fig.  ]  05  is  a  modification  of  Fig.  104.  Steel  tables  rivetted  to 
the  plate  are  substituted  for  the  lugs  used  in  Fig.  104.  The  forge 
work  is  less;  any  number  of  tables  may  be  used;  and  the  main 
plate  may  be  reduced  to  a  thickness  determined  by  considera- 
tion of  shear  and  tension  alone.  In  Fig.  104  the  plate  thickness 
is  determined  by  the  thickness  required  of  the  lug  to  prevent  it 
straightening  under  load.  No  table  should  be  placed  under  the 
foot  of  the  batter  post,  for  the  seat  for  the  table  is  usually  cut  a 
little  deeper  than  the  table,  so  the  full  bearing  area  under  the 
post  will  not  be  obtained  with  a  table  under  it. 

The  computations  for  Fig.  105  are  as  follows: 

Depth  of  toe  as  in  Fig.  104,  4  in. 

Area  required  for  bearing  between  upper  and  lower  chord 
28,125 


A  10-in  depth  will,  therefore,  be  required  for  the  upper  chord, 
giving  an  area  of  8  X  13  ins.  =  104  sq.  ins. 


180  PRACTICAL  STRUCTURAL  DESIGN 

4Q  000 
Depth  of  tables  (assuming  3  used)  =  =--       ^— -  =  1.275  ins. 

o  X  o  X  lOUU 

Use  1T58  X  3  ins. 
Assuming  three  rivets  in  each  table,  stress  in  each  rivet  = 

49  000 

— ^ —  =  5450  Ibs.    Use  three  f-in.  rivets  in  each  table. 

Thickness  of  plate  for  bearing  against  rivets  =  f  in. 

49  000 
Thickness  of  plate  for  shear  =         '  — -  =  0.614  in. 

1U,UUU  X  o 

49  000 
Thickness  of  plate  for  tension  =  36>000  x  (8'm.  _  2.8in0  -  0.59 

in.     Make  plate  f  in.  thick. 

, ,               ,      .    ..        i  .  , .         1.3125  in.  +  0.625  in.     49,000 
Moment  of  rotation  of  tables  =  5 X  — ^ — 

=  15,800  in.  Ibs. 

i  ^  soo 
Stress  in  bolts  =  ^^  =  4520  Ibs. 

,  o.O 

Add  stress  due  to  pin  in  bolster  =  |  X  ?  X  800  Ibs.  X  8  ins.  = 
800  Ibs. 

Total  stress  in  two  bolts  =  5320  Ibs.    Use  two  f-in.  bolts. 

Using  two  f-in.  diagonal  bolts,  the  horizontal  component  in 
the  bolster  will  be  as  in  Fig.  104,  requiring  one  pin. 

49  000 

Distance  required  between  tables  for  shear  =  = ~ — -==  = 

o  X  o  X  lou 

13.6  ins.    Usel3fins. 
Fig.  106  illustrates  a  practically  perfect  joint,  except  that  it 

is  not  cheap.    The  chord  and  batter  post  stresses  are  transmitted 

to  the  gusset  plates  by 
means  of  lag  screws  acting 
in  shear.  There  is  no  ec- 
centricity of  stresses  and 
consequently  no  secondary 
stresses.  With  good  in- 
spection the  lag  screws  will 
fit  closely.  The  holes  in  the 
R|  •  13 Spaces® 3"-j '-3"  —*fi'  steel  plates  for  the  lag 

•  Holes  in  plates  for  lag  screws  drilled  hWdiam     Screws     are     drilled     ^     in. 
Black  spots  sf,o«  lag  screws  in  near  plate.          fa  diameter     than    the 

jOpcn  circles  snow  lag  screws  in  far  plate. 

diameter  of  the  screws.   The 

plates  should  be  fitted  to 

the  timber  and  the  holes  marked,  after  which  they  are  bored 


JOINTS   AND  CONNECTIONS  181 

in  the  wood  to  a  diameter  a  trifle  less  than  the  diameter  of  the 
shank  of  the  screws  at  the  base  of  the  threads.  The  lag  screws 
are  to  be  screwed,  not  driven,  into  place.  Bolts  should  not  be 
used,  for  it  would  be  next  to  impossible  to  fit  the  plates  so  the 
holes  will  all  be  in  line.  To  make  a  fit  the  bolts  would  require 
some  bending  and  thus  much  of  their  value  in  shear  would  be  lost. 

"This  type  of  end  detail  is  well  suited  to  t'russes  of  an  A 
shape,  resting  upon  posts.  The  side  plates  in  such  cases  may  be 
extended  to  engage  the  top  of  the  post,  and  thus  to  give  consid- 
erable stiffness  to  the  building  frame."  —  Dewell. 

Computations  for  Fig.  106. 

56  500 

Number  of  lag  screws  in  upper  chord  =      '        =  47. 

,      ,     49,000 
Number  of  lag  screws  in  lower  chord  =     '        =  41. 

Thickness  of  plate  =  fV  m- 

Intermediate  Joints  in  Trusses 

Intermediate  joints  in  trusses  must  follow  the  general  rule  for 
joints  in  wood,  that  the  carpenter  work  must  be  as  simple  as  possible. 

The  condition  must  be  satisfied  that  the  center  lines  of  all 
members  must  meet  at  a  common  point.  In  nearly  all  joints  of 
the  types  shown  in  Fig.  107  and  Fig.  108  it  often  happens  that 
when  all  the  center  lines 
meet  at  a  common  point 
the  hole  for  the  rod  will  cut 
away  a  part  of  the  strut, 
or  the  toes  of  the  struts 
will  bear  against  the  rods. 
Sometimes  this  condition 
cannot  be  avoided  if  the 
strut  is  to  be  dapped  into  Fis-  107- 

the  chord.  Quoting  again  from  Dewell:  "If  it  so  happens  that 
the  rod  has  not  a  driving  fit  in  the  chord,  which  condition  will 
usually  exist,  especially  with  an  upset  rod  and  a  deep  chord,  the 
toe  of  the  strut  will  have  bearing  against  the  chord  for  only  part 
of  its  width.  The  result  of  this  condition  will  be  that  the  actual 
bearing  area  may  not  be  over  one-half  of  what  was  assumed  in 
design,  and  the  unit-bearing  stress  may  consequently  be  double 
the  allowable." 


182  PRACTICAL  STRUCTURAL  DESIGN 

Two  methods  of  framing  intermediate  joints  are  shown  in  Fig. 
107  and  Fig.  108.  They  are  very  common  and  yet  violate  the 
principle  that  the  carpenter  work  should  be  simple.  When  the 
strut  is  not  normal  to  the  member  it  abuts  against,  the  two  sur- 
faces of  the  indent  must  be  separately  investigated  and  the  bear- 
ing pressure  found,  for  each.  The  unit  bearing  pressure  having 
been  found  the  "minimum  bearing  area  must  then  be  determined 
by  methods  already  given.  It  involves  considerable  "cut  and 
try  "  work.  It  is  also  imperative  that  the  exact  angles  used  must 
be  marked  on  the  drawings  so  the  carpenters  can  make  the  joints 
in  the  field  and  secure  the  conditions  assumed  in  the  design.  The 
angles  of  cuts  having  been  found  so  the  bearing  is  correct  on  each 
face,  the  depth  of  each  cut  is  fixed  by  the  bearing  stress  on  the 
ends  of  the  fibers,  at  the  assumed  angles.  In  Fig.  107  the  cuts 
are  not  normal,  the  stress  actually  acting  along  the  center  line 
of  the  strut,  or  so  nearly  along  the  center  line  that  the  moment 
due  to  eccentricity  may  be  neglected.  In  Fig.  108  the  cuts  are 

normal  and  the  total  thrust 
is  assumed  to  act  over  the 
face  of  the  normal  cut.  The 
unit  stress  on  the  fibers  of  the 
chord  is  found  as  shown  hi 
the  design  of  the  cut  for  the 
end  brace.  The  depth  of  the 
cut  is  then  found.  At  the 
upper  end  the  normal  (right 
angle)  cut  is  on  the  lower 
side  of  the  strut  and  at  the  lower  end  it  is  on  the  upper  side. 
Draw  lines  through  the  centers  of  these  normal  areas,  parallel 
with  the  top  and  bottom  of  the  strut.  The  eccentricity  is  the 
distance  between  these  center  lines. 

Multiply  the  thrust  by  the  eccentricity  in  inches  and  get  the 
bending  moment  in  inch  pounds.  This  has  a  tendency  to  make 
the  end  of  the  strut  move  on  the  face  of  the  normal  cut  and 
"jump  out."  It  must  be  resisted  by  the  friction  of  the  wood  on 
the  face  of  the  cut.  Divide  the  eccentric  moment  by  the  length 
of  the  strut  in  inches  and  this  gives  the  force  to  be  developed  by 
friction.  Assuming  the  coefficient  of  wood  against  wood,  for 
sliding  friction,  to  be  0.2,  multiply  the  direct  thrust  by  0.2  and 
obtain  the  resistance  the  wood  will  offer  against  being  forced  out 


JOINTS   AND   CONNECTIONS  183 

of  the  cut  by  the  bending  moment.  Nails  and  spikes  offer  resist- 
ance against  being  pulled  out,  so  if  the  ends  of  the  strut  are  "toe 
nailed  "  this  additional  resistance  will  be  good.  It  seldom  happens 
that  the  eccentric  moment  divided  by  the  length  of  the  strut  will 
give  an  amount  exceeding  the  direct  thrust  multiplied  by  the  co- 
efficient of  friction,  but  if  it  does  then  spikes  or  bolts  must  be 
used  to  hold  the  toe  in  place.  Carelessness  in  keeping  all  joints 
tight  reduces  the  effect  of  friction,  and  decay  in  the  joint  also 
seriously  affects  it.  In  the  j  obits  illustrated  there  is  often  a 
serious  loss  in  the  efficiency  of  the  upper  and  lower  chords  because 
of  the  depth  of  the  indent.  Details  tending  to  reduce  cutting 
into  chords  should  be  favored. 

It  has  been  said  that  all  forces  should  act  through  the  center 
lines  of  members.  All  the  detailing  is  done  with  this  in  mind. 
Due  to  careless  detailing,  or,  if  the  detailing  has  been  good,  then 
due  to  careless  framing,  any  variation  in  the  relation  of  web  mem- 
bers meeting  in  a  panel  point  may  increase  secondary  stresses  to 
a  dangerous  amount.  The  horizontal  component  of  the  diagonal 
thrust  acts  through  the  lower  chord  on  a  line  intersecting  the 
center  of  bearing  of  the  thrust.  The  tension  in  the  chord  acts  on 
the  center  line  through  the 
uncut  portion  of  the  chord. 
There  is  a  moment  devel- 
oped  by  the  vertical  dis- 
tance between  these  two 

lines  of  action.    The  ver-  Vs^£~ — \^f]  Dar%" 

tical  component  of  the 
thrust  acts  through  the 
center  of  the  face  of  the 
cut.  This  forms  a  couple  Fis-  109' 

with  the  tension  in  the  vertical  rod.  There  is  a  moment  developed 
by  the  distance  between  the  center  of  the  rod  and  the  line  of  action 
of  the  vertical  component  of  the  thrust  in  the  strut.  In  wooden 
trusses  the  secondary  stresses  are  seldom  important  except  when 
very  high  unit  stresses  are  used,  but  we  cannot  afford  to  neglect 
consideration  of  the  possibility  of  neglect  to  keep  the  joints  tight 
and  the  possibility  of  rotting  setting  in.  It  is  very  little  trouble 
to  investigate  the  effect  of  secondary  stresses  and  provide  addi- 
tional material,  or  to  redesign  the  joint  to  reduce  the  secondary 
stresses  to  a  minimum.  It  is  at  least  important  that  secondary 


184  PEACTICAL  STRUCTURAL  DESIGN 

stresses   should   be   investigated   when   high   unit   stresses   are 
used. 

The  advantages  of  the  type  of  joint  shown  in  Fig.  109  are  best 
summed  up  in  the  words  of  Mr.  Dewell:1  "In  this  joint,  the 
strut  has  a  full  bearing  on  the  butt  block,  and  the  butt  block, 
in  turn,  utilizes  the  total  width  of  the  chord  for  bearing.  Also, 
the  detail  takes  advantage  of  the  full  bearing  pressure  in  end  com- 
pression of  the  butt  block  on  the  chord,  resulting  in  a  minimum 
depth  of  cut  into  the  chord.  Nearly  all  the  cuts  are  normal, 
and  the  others  are  simple.  All  the  cuts  can  be  easily  and  accu- 
rately laid  out  and  made  by  the  carpenter.  The  length  of  the 
butt  block  can  be  adjusted  to  fit  all  conditions  of  possible  inter- 
ference with  other  connections.  Its  minimum  length  is  deter- 
mined by  longitudinal  shear.  The  bolt  through  the  end  of  the 
butt  block  holds  the  block  securely  in  its  socket.  Whether  there 
is  any  actual  tension  in  the  bolt  depends  upon  the  length  of  the 
butt  block.  This  can  be  determined  at  once  by  inspection.  If 
the  line  of  the  thrust  of  the  strut  falls  within  the  base  of  the 
block,  there  can  be  no  tension  in  the  joint.  However,  it  is  well 
to  provide  at  least  a  f-in.  bolt  to  bind  the  joint  together  thor- 
oughly." In  another  place,  Mr.  Dewell  says:  "The  detail  of 
Fig.  109  is  seldom  used;  nevertheless  it  is  the  most  consistent 
and  logical  in  principle  and  the  simplest  of  construction  of  the 
three  types  shown."  In  the  foregoing  remarks  the  author 
heartily  concurs.  In  too  many  cases  draftsmen,  not  entitled 
to  be  termed  designers,  merely  butt  opposing  diagonals  against 
one  another  with  no  provision  for  transmitting  the  component 
of  the  diagonal  stress  to  the  chord.  Designers  must  never  forget 
that  all  forces  can  be  assumed  to  act  along  lines:  these  lines  in- 
tersect lines  in  other  members  and  the  force  is  then  divided  and 
goes  in  two  directions.  The  main  force  is  termed  the  resultant 
and  the  other  forces  the  components.  This  will  be  discussed 
fully  in  the  chapter  on  Graphic  Statics. 

Pin  Connections 

A  pin  connection  is  sometimes  an  economical  connection.  It  may 
be  used  with  either  wooden  or  metal  frames.  The  pieces  connected 

1  Western  Engineering,  Oct.  1916,  p.  386. 


JOINTS  AND  CONNECTIONS 


185 


by  the  pin  must  have  enough  bearing  area  to  prevent  crushing. 
This  being  attained  the  pin  is  designed  to  resist  shear  and  bending. 

Pin  connections  require  a  minimum  of  material  in  the  members. 
The  cost  of  fabrication  with  pin  connected  trusses  is  not  high. 
Such  trusses  cost  less  than  rivetted  trusses.  The  joint  is  flexible, 
if  the  pin  does  not  rust,  for  all  forces  meet  on  the  axis  of  the  pin. 
It  is  theoretically  a  perfect  joint  and  for  many  years  was  favored 
by  American  bridge  engineers  for  the  reasons  given.  European 
engineers  always  favored  the  rivetted  joint  because  of  its  rigidity 
and  all  joints  were  designed  to  take  care  of  eccentric  stresses. 
Under  heavy  traffic  it  was  found  that  the  pin  holes  wore  badly 
and  thus  the  trusses  became  too  flexible.  When  pins  rusted  into 
place  eccentric  stresses  were  set  up  and  frequently  the  members 
were  too  small  to  take  care  of  them.  The  pin-connected  joint 
at  the  present  time  is  not  high  in  favor  with  bridge  engineers,  but 
it  is  all  right  for  roof  trusses. 

Referring  to  Fig.  110  the  bearing  area  is  found  by  assuming 
the  whole  load  to  rest  on  a  strip  having  a  length  equal  to  the 
combined  thicknesses  of  the  pieces 
connected,  with  a  width  equal  to 
the  diameter  of  the  pin.  Some- 
times, for  example  in  the  case  of 
a  built-up  member,  an  extra  thick- 
ness of  steel  is  rivetted  to  the 
side  of  a  member  in  order  to  ob- 
tain increased  bearing  area  at  the 
pin  hole.  This  is  often  cheaper 
than  to  increase  the  thickness  of 
the  metal  in  the  member  through- 
out the  whole  length. 


Fig.  110. 


The  shear  on  the  pin  seldom  determines  the  thickness,  the  bear- 
ing area  and  bending  stresses  being  usually  of  greater  importance. 
The  shear,  however,  should  in  all  cases  be  investigated.  The  num- 
ber of  joints,  that  is  the  divisions  between  the  pieces,  will  be  one 
less  than  the  number  of  pieces.  Divide  the  sum  of  the  loads 
on  the  pin  by  the  number  of  joints  to  obtain  the  shear  on  each 
joint,  if  the  pieces  are  of  equal  thickness.  If  they  are  not  equally 
thick  the  shear  on  each  joint  will  be  equal  to  the  sum  of  the  re- 
actions of  the  pieces  on  either  side  of  the  joint.  If  the  pin  is 
found  to  be  too  small  to  carry  the  shear  the  diameter  must  be 


186  PRACTICAL  STRUCTURAL  DESIGN 

increased,  which  will  have  the  effect  of  reducing  the  unit  stress  in 
bearing. 

To  determine  the  flexure  hi  pins  the  following  formula  is  used 
for  the  resisting  moment: 


32        8 

in  which 

M  =  moment  of  forces  for  any  section  through  the  pin. 
/  =  allowable  unit  fiber  stress  in  bending. 
TT  =  3.1416. 
d  =  diameter  of  pin. 
A  =  cross-sectional  area  of  pin. 

The  load  in  every  member  must  be  reduced  to  the  horizontal 
and  vertical  component  loads,  and  must  be  considered  as  acting 
in  each  member  along  the  center  line,  so  that  the  point  of  applica- 
tion of  each  horizontal  and  vertical  component  is  at  the  center 
of  bearing  of  the  corresponding  member.  This  means  that  if 
two  |  -in.  bars  are  side  by  side  the  moment  arm  =  \  +  \  =  \  in. 
The  horizontal  forces  are  equal  on  both  sides  of  the  pin,  other- 
wise there  would  not  be  equilibrium.  Similarly  the  vertical  force 
downward  is  equal  to  the  upward  acting  vertical  force. 

The  bending  moment  (to  which  the  resisting  moment  must  be 
equated)  is  as  follows  : 

M  =  V(Mh)2  +  (Mv)2. 

in  which  M  =  resulting  bending  moment  in  inch  pounds. 
(Mh)  =  maximum  moment  of  all  horizontal  stresses. 
(Mv)  =  maximum  moment  of  all  vertical  stresses. 
In  designing  pin  joints  no  two  adjacent  bars  should  pull  in  the 
same  direction,  unless  they  shall  by  so  doing  reduce  the  bending 
moment.     The  joint  must  be  symmetrically  arranged  to  avoid 
torsion.     Diagonal  ties  should  be  placed  close  to  the  vertical 
member  and  the  horizontal  ties  should  preferably  be  on  the  out- 
side.   Sometimes  packing  pieces  are  required  between  the  mem- 
bers that  carry  stress,  but  these  packing  pieces  merely  lengthen 
the  moment  arm  between  adjacent  members.     The  joint  being 
symmetrical  the  computation  stops  at  the  center  piece. 

In  determining  the  horizontal  moments  take  one-half  the  sum 
of  the  thickness  of  adjacent  bars  for  the  moment  arm  between 
these  two  bars.  The  moment  between  the  first  two  bars  is  equal 
to  the  load  on  the  outer  bar  times  the  moment  arm.  The  moment 


JOINTS   AND   CONNECTIONS 


187 


between  the  second  and  third  bars  is  equal  to  the  moment  just 
found  plus  the  difference  between  the  loads  on  the  first  and 
second  bars  times  the  moment  arm  between  the  second  and  third. 
The  moment  between  the  third  and  fourth  bars  is  equal  to  the 
last  moment  plus  the  difference  between  the  load  on  the  first  bar, 
less  the  sum  of  the  loads  on  the  second  and  third,  times  the  mo- 
ment arm  between  the  third  and  fourth  bars.  In  determining 
the  vertical  moment  multiply  the  vertical  load  by  the  distance 
between  centers  of  the  vertical  member  and  the  most  distant 
inclined  member.  If  there  are  a  number  of  inclined  members 
then  proceed  as  in  computing  the  horizontal  moments,  using 
the  vertical  loads. 

Referring  to  Fig.  110  the  loads  on  the  members  are  designated 
as  PI,  P2,  P3  and  P4.  The  moment  arms  are  designated  as  A, 
B,  C,  and  D,  being  in  inches.  The  resulting  moments  are  desig- 
nated by  Ma,  Mb,  Mc,  and  M*. 

At  P2  the  moment  =  Ma  =  PI  x  A. 

At  P3  the  moment  =  M b  =  Ma  +  (Pi  -  P2)  X  B. 

At  P4  the  moment  =  Mc  =  Mb  +  (Pi  -  P2  +  PS)  X  C. 

The  members  P4  are  inclined  and  the  center  piece  is  vertical. 
The  vertical  moment  is  equal  to  the  vertical  load  multiplied  by 
the  arm  D.  The  vertical  member  is  made  of  two  channels  and  the 
other  members  are  eye  bars. 

Fig.  Ill  is  from  the  1913  edition  of  the  Carnegie  Pocket  Com- 
panion.   A  pin  has  to  carry  a  load  of  64,000  Ibs. : 
required  the  size  at  24,000  Ibs.  fiber  stress,  assuming 
the  distance  between  points  of  support  to  be  5  ins. 

Bending  moment  =  64,000  x  5  -=-  4  =  80,000  in. 
Ibs.  This  it  is  seen  considers  the  center  load  as 
concentrated  and  allows  nothing  for  the  distribu- 
tion of  the  load  over  a  part  of  the  span.  The  size 
of  the  pin  may  be  obtained  from  the  table  on  page 
219  in  that  book.  Looking  in  the  column  headed 
by  24,000  find  the  nearest  (larger)  resisting  moment, 
which  is  80,900  in.  Ibs.  In  the  first  column  at  the 
left  is  the  diameter,  3|  in. 

The  size  and  diameter  of  the  pin  may  also  be  found  from  the 
expression,  M  =  fAd  -f-  8. 


188  PRACTICAL  STRUCTURAL  DESIGN 

First  divide  /  by  8  =  24,000  -r  8  =  3000. 
Then  M  =  3000  Ad 


0.7854  d2,  therefore  Ad  =  0.7854  d3. 

80,000 
0.7854  x  3000' 


-4 


80,000 


0.7854  x  3000 

Many  tension  members  in  steel  work  are  made  of  round  rods 
or  rectangular  eye  bars.  The  ends  are  fastened  to  the  frame  by 
means  of  pins  passing  through  loops  or  yokes  or  eyes.  The  area 
of  the  main  part  of  the  member  is  found  by  dividing  the  total 
tension  by  the  allowable  fiber  stress.  The  thickness  of  the  loop, 
the  yoke  or  the  eye  is  determined  by  the  required  bearing  area 
on  the  pin.  If  the  enlarged  section  on  the  end  to  receive  the  pin 
is  welded  to  the  member  the  stress  used  should  be  low  to  allow 
for  imperfections  in  the  welding.  If  the  members  are  purchased 
from  the  mills  already  welded  they  should  be  purchased  under 
very  rigid  specifications.  The  use  of  clevises,  turnbuckles  and 
sleeve  nuts  permits  tension  members  to  be  lengthened  and 
adjusted  for  length. 

In  the  Carnegie  Pocket  Companion,  1913  edition,  all  the  in- 
formation the  designer  needs  about  the  sizes  of  screw  threads, 
bolts,  eye  bars,  loop  rods,  clevises,  turnbuckles  and  sleeve  nuts 
is  found  on  pages  112  to  122  inclusive,  215,  218,  219,  223. 

Similar  information  is  found  on  pages  322  and  pages  331  to 
357  inclusive  in  the  1914  edition  of  the  Cambria  Steel  Hand  Book. 
In  the  1916  edition  of  Jones  &  Laughlin,  Standard  Steel  Con- 
struction, this  information  is  on  page  246  and  on  pages  255  to  268 
inclusive.  The  Lackawanna  Steel  Company  Hand  Book  contains 
similar  information  on  pages  339  to  363  inclusive. 

Rivets  and  Rivetting 

A  rivet  is  a  piece  of  metal  which  connects  together  two  or 
more  pieces  of  metal.  In  structural  work  rivets  are  made  of  soft 
steel.  A  head  is  formed  on  one  end  of  a  rivet  when  it  is  made  and 
when  used  the  rivet  is  heated  and  the  surplus  length  projecting 


JOINTS   AND   CONNECTIONS  189 

beyond  the  plates  is  formed  into  a  head  by  means  of  presses  in 
the  shop  or  by  hammers  in  the  field. 

A  rivet  should  never  be  used  in  tension  when  it  is  possible  to 
avoid  so  using  it.  In  very  exceptional  cases  a  rivet  may  have  to 
be  so  used  and  then  the  allowable  tensile  stress  should  not  exceed 
8000  Ibs.  per  sq.  in.  The  body  of  the  rivet  when  used  in  tension 
may  be  amply  large,  but  the  thickness  of  the  head  must  be  inves- 
tigated to  determine  whether  it  will  be  sheared  by  the  pull  on 
the  rivet,  this  shearing  being  on  a  circle  having  a  diameter  equal 
to  the  diameter  of  the  body  of  the  rivet.  The  head  of  the  rivet 
must  be  thick  enough  to  withstand  the  shear. 

The  reason  for  not  using  rivets  in  tension  is  that  the  unequal 
heating  and  cooling  during  the  process  of  fabrication  of  a  member 
which  is  rivetted  together  sets  up  in  the  rivet  expansion  and  con- 
traction stresses  of  unknown  amount.  It  has  happened  many 
times  that  rivet  heads  were  snapped  off  when  cooling  and  they 
snap  off  sometimes  in  extremely  hot  weather  and  in  extremely 
cold  weather.  With  such  stresses  existing  in  rivets  it  is  mani- 
festly dangerous  to  further  impose  on  them  a  direct  tensile  stress, 
for  the  rivet  heads  may  be  on  the  verge  of  snapping  off  and  any 
slight  additional  load  may  cause  them  to  go.  It  is  best  to  use 
bolts  in  joints  in  which  rivets  would  be  subjected  to  tension  when 
field  rivets  are  driven  and  if  shop  driven  rivets  are  under  tension 
the  stress  should  be  very  low.  Shop  driven  rivets  are  pressed 
into  place  and  all  the  conditions  in  the  shop  make  it  likely  that 
the  work  is  uniform.  It  is  impossible,  however,  to  secure  proper 
conditions  in  the  field,  for  the  heat  cannot  be  controlled  and  many 
rivets  are  burned.  They  are  thrown  through  the  air  and  driven 
after  some  cooling  has  taken  place.  The  hammering  may  be 
uneven  and  the  rivets  may  not  be  hot  enough  when  driven  to 
be  forced  to  completely  fill  the  hole. 

Rivets  are  assumed  to  act  entirely  in  shear  and  all  computa- 
tions for  rivetted  joints  are  based  on  this  assumption.  There 
can  be  no  doubt  that  friction  is  a  big  factor  in  rivetted  joints, 
the  rivets  in  shrinking  drawing  the  plates  together  and  holding 
them  in  contact  so  that  friction  between  the  plates  assists  the 
shearing  strength  of  the  rivets.  The  assistance  obtained  from 
friction  is  neglected  in  computations  and  merely  increases  the 
factor  of  safety  of  the  joints. 

Rivets  may  fail  by  bending.     This  effect,  however,  is  not 


190 


PRACTICAL  STRUCTURAL  DESIGN 


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oo~  oo- 


t>-       1-1 

of      e*f 


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O    I> 

co  of 


O  O 
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l>  00         O5  ^H 


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00 


JOINTS   AND  CONNECTIONS  191 

important  except  in  very  long  rivets  holding  several  plates.  The 
action  is  then  similar  to  that  on  a  pin  and  is  investigated  simi- 
larly. It  may  be  advisable  then  to  use  bolts  or  pins. 

Rivets  are  much  cheaper  than  bolts,  otherwise  bolts  would 
be  used,  for  they  are  safe  in  bending  and  in  tension  as  well 
as  in  shear.  An  objection,  however,  to  bolts  is  that  it  is 
difficult  to  screw  the  nuts  tight  and  keep  them  from  becoming 
loose  under  vibration.  When  bolts  are  used  where  rivets,  if 
used,  might  be  in  tension,  means  must  be  provided  for  keeping 
the  nuts  tight.  Even  the  best  nut  locks  require  frequent 
inspection.  •  * 

The  accompanying  table  from  the  Jones  &  Laughlin  Hand 
Book  gives  the  value  of  rivets  in  plates  of  different  thicknesses. 
The  values  used  are  common  and  the  steel  handbooks  all  contain 
tables  for  other  values. 

In  single  shear  rivets  connect  two  plates,  so  there  is  one  joint 
on  which  the  plates  may  slide,  precisely  like  the  blades  of  shears. 
In  double  shear  rivets  connect  together  three,  or  more,  plates 
so  there  are  at  least  two  joints  on  which  the  plates  may  slide. 
When  three  plates  are  used  the  middle  plate  is  assumed  to  be 
pulling  out  from  between  the  two  outer  plates. 

In  the  table  the  thickness  of  the  plates  connected  by  rivets  is 
given  together  with  the  bearing  value  and  the  shearing  value. 
Figuring  shear  at  10,000  Ibs.  per  sq.  in.  the  value  of  the  rivet  in 
shear  is  constant,  no  matter  what  the  thickness  of  the  plate. 
The  bearing  value  of  the  plate  is  20,000  Ibs.  per  sq.  in.  and  a 
comparison  of  the  figures  shows  immediately  when  the  value  of 
a  rivet  is  determined  by  bearing  and  when  by  shear. 

The  strength  in  double  shear  is  of  course  just  twice  that  in 
single  shear.  The  bearing  on  the  plate  is  determined  by  the 
thickness  of  a  single  plate,  of  two  adjacent  plates  in  which  the 
stresses  are  opposite.  If  two  or  more  adjacent  plates  are  fastened 
together  and  act  as  one  plate,  then  the  plate  thickness  is  the 
combined  thickness  of  the  plates. 

In  the  table  the  strength  of  the  ri vetted  joint  is  determined  by 
the  shearing  value  of  the  rivet  above  the  heavy  line  in  some  of 
the  columns.  Below  the  heavy  lines  the  strength  is  fixed  by  the 
bearing  value  of  the  plate. 

To  determine  the  number  of  rivets  to  use  first  select  the  size  of 
rivet.  The  thickness  of  the  plate  is  determined  when  designing 


192  PRACTICAL  STRUCTURAL  DESIGN 

the  member.     Looking  in  the  table,  in  the  column  headed  by 
thickness  of  plate,  opposite  the  size  of  rivet  will  be  found  the 
bearing  value.    In  the  column  headed  by  shear  find  the  shearing 
'^W//M^MM^/M  value  of  the  rivet.    Use  the  smaller 

^\^m\^s^  value.    Divide  the  total  load  by 

?      |~~^      ^~]       ^j        tn^s  selected  value  and  obtain  the 
*•  —  !  —  -  -  -  —  I—  -  *        number  of  rivets.     If  the   rivets 
are   in   double  shear   double  the 
/i    shearing  value  given  in  the  table 
and  compare  it  with  the  bearing 
ji  ^      _    I    <      value,  using  the  smaller  amount. 

The  rivets  girted  in  this  way  are 


>  I    XT 

{  1 


Double  Shear  g^fe  against  destruction  by  shear- 

Fig.  112.  ^g  across  an(j  the  edges  of  the 

holes  through  the  plate  will  not  crush.    Fig.  112  illustrates  single 
and  double  shear. 

Let  /  =  bearing  value  on  plate  in  Ibs.  per  sq.  in. 
t  =  thickness  of  plate  in  inches. 
v  =  shearing  value  of  rivet  in  Ibs.  per  sq.  in. 
d  =  diameter  of  rivet. 
Then 

Shearing  value  in  single  shear  =  0.7854  x  d2  x  v. 

Shearing  value  in  double  shear  =  2  x  0.7854  x  d2  X  V. 

Bearing  value  =  d  x  t  x  /. 

When  driven,  rivets  are  assumed  to  completely  fill  the  holes, 
and,  therefore,  in  compression  pieces  no  reduction  in  area  is  made 
for  the  rivet  holes.  Tensile  stress  cannot,  however,  be  trans- 
mitted through  the  rivets  and  the  area  occupied  by  rivet  holes 
weakens  the  piece  in  tension.  If  the  member  is  narrow  and  one 
hole  is  drilled,  or  punched  through  it,  the  piece  must  be  increased  in 
width  or  thickness  to  make  up  for  the  area  removed  by  the  hole. 
Thus,  if  in  a  bar  4  ins.  wide  and  f  in.  thick  a  hole  f  in.  diameter 
is  made,  the  bar  will  be  increased  in  width  by  f  in.,  or  the  thick- 
ness will  be  increased,  provided  the  width  must  be  maintained. 
The  thickness  is  increased  as  follows:  The  width  left  is  3.25  ins. 
The  area  removed  is  \  X  f  =  f  sq.  in.  The  thickness  to  be  added 
equals  0.3750  -f-  3.25  =  Tft-  in.,  making  the  total  thickness  \\  in. 
There  may  be  one  hole  through  the  plate  or  there  may  be  several. 
If  the  holes  are  in  line  the  cross-sectional  area  of  one  hole  is  de- 
ducted. If  the  holes  are  in  two  lines  the  cross-sectional  area  of 


JOINTS   AND   CONNECTIONS  193 

two  holes  will  be  deducted.  Similarly  three  holes  will  be  deducted 
for  three  lines,  etc.  It  means  that  a  strip  of  metal  equal  to  the 
width  of  the  rivet  hole  is  ignored  provided  all  the  rivets  are 
driven  within  the  strip,  or  if  the  rivets  are  driven  in  two  or  more 
strips  then  these  strips  are  ignored  in  making  computation  for 
strength  in  tension. 

A  structural  designer  will  determine  the  number  of  rivets  in 
the  manner  described.     He  may  then  arrange  them  in  a  group 


O  O  i'O  O 
OOSOO 
O  O  i  O  O 


Fig.  113.  Fig.  114. 

similar  to  that  shown  in  Fig.  113,  for  structural  designers  do  not 
always  study  joint  efficiency.  A  boiler  maker,  on  the  contrary, 
would  study  the  joint  in  order  to  obtain  the  maximum  efficiency 
and  his  detail  would  resemble  Fig.  114.  There  is  no  reason  why 
a  structural  designer  should  not  obtain  the  highest  possible 
efficiency  in  designing  rivetted  joints. 

Assume  that  the  tension  in  the  joint  is  30,000  Ibs.  and  each 
rivet  carries  5000  Ibs.  In  Fig.  113  three  rivet  holes  must  be  de- 
ducted, which  means  additional  material  added  to  the  member, 
for  only  the  material  left  after  the  holes  are  made  can  carry  ten- 
sion. The  rivets  carry  shear  at  an  assumed  unit  shearing  stress. 
The  holes,  however,  cut  out  an  area  which  carries  tension  at  an 
assumed  unit  tensile  stress. 

In  Fig.  114  the  end  rivet  carries  one-sixth  of  the  stress  as 
shear.  The  plate  on  a  section  through  the  hole  carries  the  total 
stress.  The  diameter,  therefore,  of  the  rivet  hole  is  one-seventh 
of  the  total  width  of  the  plate.  On  a  section  through  the  two 
rivets  the  plate  loses  in  each  hole  one-seventh vpfrthe  total  width, 
but  as  it  had  enough  area  to  carry  seven-sixths  "of  the  load  the 
cutting  of  two  holes  leaves  enough  area  to  carry  five-sevenths  of 
the  load. 

The  plate  with  an  area  sufficient  to  carry  five-sevenths  of  the 
load  then  reaches  the  section  near  the  end  where  there  are  three 
rivet  holes.  Each  hole  has  a  diameter  of  one-seventh  the  width 
of  the  plate  and  the  three  holes  leave  four-sevenths  of  the  area 
intact. 

At  the  single  rivet  one-sixth  of  the  stress  is  carried  in 


194  PRACTICAL  STRUCTURAL  DESIGN 

while  the  plate  can  carry  the  whole.  At  the  line  of  the  two  rivets 
one-half  (three-sixths)  of  the  stress  is  carried  by  shear  while  the 
remainder  of  the  plate  can  carry  five-sevenths  of  the  load.  At 
the  line  of  the  three  rivets  all  the  stress  can  be  carried  by  shear 
while  the  plate  has  area  enough  remaining  to  carry  four-sevenths 
of  the  load.  Therefore,  by  this  arrangement  of  rivets  it  is  neces- 
sary to  deduct  only  the  width  of  one  hole,  whereas  if  the  rivets 
are  arranged  as  in  Fig.  113  it  will  be  necessary  to  deduct  the  width 
of  three  holes. 

When  an  arrangement  is  made  such  as  that  shown  in  Fig.  114 
the  splice  plates  will  be  a  little  longer  and  possibly  a  little  thicker 
than  the  plates  used  in  an  arrangement  such  as  that  shown  in 
Fig.  113.  This  is,  however,  offset  by  the  fact  that  when  material 
is  added  to  overcome  the  cross-sectional  area  cut  out  by  holes, 
the  material  is  added  to  the  area  of  the  member  throughout 
its  whole  length.  The  saving  effected  by  a  design  like  Fig.  114 
is  two  rivet  holes  when  compared  with  the  design  shown  in 
Fig.  113. 

The  efficiency  principle  is  applicable  only  to  splices  and  in 
connections  of  rivetted  trusses.  There  are  many  cases  in  which 
the  efficiency  principle  must  be  disregarded,  except  as  it  affects 
the  sizes  of  rivets  used. 

A  rivetted  joint  fails  by  shearing  of  the  rivets  or  by  the  metal 
between  the  rivets  giving  way.  The  distance  between  centers 
of  rivets  is  termed  the  pitch.  It  is  fixed  partly  by  considering 
the  shearing  strength  of  the  metal.  It  is  fixed  partly  by  arbitrary 
specifications.  It  is  fixed  partly  by  the  requirement  that  the 
section  between  rivet  holes  must  be  fully  as  strong  as  the  rivet. 

Referring  to  the  rivets  shown  in  Fig.  113  and  Fig.  114.  The 
plate  if  it  tears  will  tear  at  the  weakest  point.  This  may  be  on 
the  vertical  line  joining  centers  of  the  rivets  and  may  be  on  a 
zigzag  line  joining  centers  of  adjacent  lines  of  rivets.  Experi- 
ments apparently  indicate  that  rupture  is  as  likely  to  occur  on 
the  zigzag  line  as  on  the  vertical  line,  the  cross-sectional  net 
area  determining  this  matter.  By  net  area  is  meant  the  area 
measured  between  edges  of  holes.  Some  specifications  require 
that  the  net  area  on  the  zigzag  line  exceed  the  square  area  by 
30  per  cent,  but  general  rules  should  never  be  followed,  except 
when  required  by  specifications.  It  is  poor  policy  for  designers 
to  follow  arbitrary  rules  when  they  are  competent  to  investigate, 


JOINTS   AND  CONNECTIONS  195 

and  have  the  time  to  investigate,  the  conditions  of  a  particular 
case.  Much  poor  designing  and  unsafe  designing  can  be  traced 
to  blind  observance  of  rules  with  apparent  disregard  of 
reasoning. 

Standards 

Fabrication  standards  are  fixed  by  the  types  and  capacities  of 
the  tools  and  machines  with  which  a  fabricating  shop  is  equipped. 
A  designer  should  know  the  standards  of  the  shop  in  which  his 
steel  work  will  be  fabricated  and  endeavor  to  arrange  his  detail- 
ing accordingly.  This  will  insure  the  lowest  possible  cost  for  his 
client. 

Standards  for  rivet  spacing  will  be  found  on  pages  212,  213, 
and  214,  Carnegie,  1913  ed. :  328,  329  and  330,  Cambria,  1914  ed.; 
246  to  259  incl.,  Jones  &  Laughlin,  1916  ed.;  336,  337  and  338, 
Lackawanna,  1915  ed.  The  student  is  advised  to  study  carefully 
a  number  of  other  pages  in  these  books,  dealing  with  the  subject 
of  rivets.  Tables  for  the  pitch  (the  distance  between  rivets)  of 
rivets  in  angles  are  based  on  the  angle  developed.  That  is,  the 
angle  bent  flat  as  if  it  were  a  narrow  plate.  This  is  an  important 
thing  to  remember  in  detailing. 

All  manufacturers  have  standard  beam  connections  based  on 
developing  the  full  strength  of  the  beam  on  the  shortest  span  on 
which  it  will  carry  the  maximum  load  without  failure  by  crip- 
pling or  shear.  When  loading  conditions  are  not  severe  some 
expense  can  be  saved  by  designing  connections  to  fit  the  case. 
When  no  details  are  shown  for  beam  connections  the  standard 
connections  are  understood.  It  is  well,  however,  to  make  a  note 
to  this  effect  on  the  drawings  and  avoid  controversy.  Refer  also 
to  the  particular  standards  wanted.  Standard  beam  connec- 
tions are  given  on  page  207,  Carnegie,  1913  ed. ;  42  to  50  incl., 
Cambria,  1914  ed.;  88,  89,  Bethlehem,  1911  ed.;  323  to  331 
incl.,  Lackawanna,  1915  ed. ;  126  to  129  incl.,  Jones  &  Laughlin, 
1916  ed. 

It  is  time  now  for  the  student  to  procure  sets  of  specifications 
for  detailing  structural  steel.  Some  of  the  handbooks  contain 
such  specifications  and  all  contain  valuable  data  for  specifications. 
The  following  specifications  are  recommended  for  purchase  and 
study : 

Standard  Specifications  for  Structural  Steel,  Timber,  Concrete 


196  PRACTICAL  STRUCTURAL  DESIGN 

and  Reinforced  Concrete.  By  John  C.  Ostrup.  Sold  by  the  U.  P.  C. 
Book  Company,  Inc.,  New  York,  for  $1.00 

General  Specifications  for  Structural  Work  of  Buildings.  By 
C.  C.  Schneider.  Sold  by  the  publisher  of  this  book  for  75  cts. 

Specifications  and  Tables  for  Steel  Framed  Structures.  Pre- 
pared by  the  American  Bridge  Company.  Distributed  by  the 
New  York  and  Chicago  offices  free  of  charge. 

Building  Code  recommended  by  The  National  Board  of  Fire 
Underwriters.  Address  the  officers  of  the  Board,  76  Williams 
Street,  New  York,  N.  Y.  Similar  information  is  to  be  had  in 
the  building  ordinances  of  all  large  cities.  Small  details,  however, 
of  the  sizes  of  members  and  spacing  of  rivets,  etc.,  can  only  be 
had  in  specifications  similar  to  the  first  three  mentioned. 

A  number  of  standard  specifications  for  structural  work  are 
sold  and  a  list  can  be  obtained  from  any  large  publishing  and 
bookselling  concern. 

Secondary  Stresses  in  Framed  Structures 

Secondary  stresses  in  framed  structures  are  due,  primarily, 
to  faulty  details.  In  the  general  design  of  a  framed  structure 
it  is  assumed  that  all  forces  meet  at  a  common  point,  which  is 
the  intersection  of  the  axes  through  the  centers  of  gravity  of  the 
members  forming  the  joint.  In  the  case  of  a  pin  connected  truss, 
with  the  pin  clean  and  the  joint  in  first-class  condition,  this 
assumption  is  very  nearly  met.  In  the  case  of  rivetted  joints 
the  direction  of  each  member  is  rigidly  fixed  and  when  the  struc- 
ture deflects  under  load  all  members  are  placed  in  double  curva- 
ture. This  condition  of  secondary  stress  is  accentuated  by  faulty 
joints,  the  resulting  stresses  with  carefully  studied  joints  being 
often  negligible.  With  faulty  joints  a  structure  may  fail  because 
of  secondary  stresses. 

The  effect  of  faulty  design  can  best  be  shown  by  an  example 
and  Fig.  115  shows  a  joint  in  the  top  chord  of  a  Warren  truss. 
Taking  A  as  the  center  of  moments  the  total  bending  moment, 
due  to  eccentricity,  is  35,600  x  7.5  =  267,000  in.  Ibs.  This  mo- 
ment is  apportioned  among  the  four  members  meeting  at  the 
joint  in  accordance  with  their  relative  rigidities,  which  is  found 
by  dividing  their  Moment  of  Inertia  (to  be  found  in  the  steel 
handbooks)  by  one-half  the  length  of  the  member,  all  in  inches. 

To  understand  this  question  of  relative  rigidity  assume  that 


JOINTS   AND   CONNECTIONS  197 

a  bending  moment  is  set  up  at  a  joint  where  all  the  members  are 
rigidly  connected  and  that  the  other  ends  of  the  members  are 
likewise  rigidly  connected.  The  members  are  bent  at  their  ends 
in  opposite  directions, 

thus  setting  up  a  double  j^^aftrofCnvify 

curvature,  with  a  point      j  \.,.?x/5^\l^6***£. 

of  contraflexure,  or 
point  of  zero  moment 
in  the  middle  of  the 
length.  The  members 
may  be  considered  as 
beams  fastened  at  the 
joint  with  the  middle 
point  the  free  end.  All 
the  members,  therefore, 
resist  the  bending  moment  in  proportion  to  their  relative  rigidities. 
The  angular  displacement  of  the  joint  is  the  same  for  all  the 
members  meeting  at  the  joint.  The  angular  displacement  at 
the  joint  then  is  the  deflection  of  the  middle  point  of  any  member 
divided  by  the  half  length  of  that  member.  This  can  be  demon- 
strated by  an  expression  in  which  appears  the  modulus  of  elas- 
ticity, the  bending  moment,  the  moment  of  inertia,  the  half  length 
and  the  angular  displacement.  Dropping  all  the  factors  common 
to  all  the  members  there  are  left  only  the  Moment  of  Inertia  and 
the  half  length,  so  it  is  readily  seen  that  the  total  bending  mo- 
ment is  divided  among  the  several  members  in  proportion  to  their 

respective  rigidities,  that  is  in  proportion  to  y    The  Moment  of 

Inertia  is  in  inches  and  the  half  length  should  properly  be  in 
inches.  However  fewer  figures  are  used  and  the  work  simplified 
by  dividing  the  Moment  of  Inertia  by  the  total  length  of  the 
member  in  feet.  The  proportionate  values  are  the  same.  This 
will  be  illustrated. 

Let  the  total  length  of  the  chord  between  joints  be  12  ft.  and 
the  depth  between  centers  of  gravity  be  6  ft.  The  length  of  the 
diagonals  will  be  8.24  ft.  but  in  the  division  we  will  use  only  8  ft. 
The  Moment  of  Inertia  of  the  top  chord  is  27,  the  value  for  one 
angle  being  13.5,  as  shown  on  page  148,  Carnegie,  1913  ed.,  and  on 
page  113,  Jones  &  Laughlin,  1916  ed.,  similar  values  being  given 
in  the  other  standard  steel  handbooks.  Similarly  the  Moment 


198  PRACTICAL  STRUCTURAL  DESIGN 

27  6  8 

of  Inertia  for  the  braces  is  6.8.    Dividing,  -=  =  2.25,  and  -—•  = 

0.85.  Suppose  the  8  and  the  12  are  divided  by  2,  to  get  the  half- 
length,  and  then  multiplied  by  12  to  reduce  the  half-length  to 
inches,  it  is  plain  to  see  that  this  is  equivalent  to  multiplying  the 
8  and  12  by  6,  so  the  result  of  the  division  in  each  case  will  be 
one-sixth  as  large  as  before,  but  the  proportion  is  unaltered. 
This  illustration  has  been  worked  out  because  it  explains  the 
appearance  of  many  expressions  which  often  cause  the  amateur 
considerable  trouble.  A  man  accustomed  to  reasoning  as  he 
figures  will  often  introduce  many  short  cuts  into  formulas  and 
expressions  which  he  alone  will  understand,  but  the  reasons  for 
which  can  be  readily  worked  out  by  any  other  equally  competent 
man. 

Returning  to  Fig.  115,  add  together  the  values  of  y  for  the 

two  chord  members  and  the  two  web  members,  the  sum  being 
(2  x  2.25)  -I-  (2  x  0.85)  =  6.19.  The  moments  are  now  distributed 
as  follows: 

2.25  x  267,000 


6.19 


=  97,000  in.  Ibs.  for  the  chord. 


0.85  x  267,000      0_  _„  •     «      ,     .  u        u          u 
-  «To    -  =  36,700  in.  Ibs.  for  the  web  members. 

The  maximum  fiber  stress  in  the  members  induced  by  these 
moments  is  as  follows  : 


For  chords,  /  =         =      >        xi  =  14,400  Ibs.  per  sq.  in.     The 

4  in.  is  the  distance  from  the  top  of  the  angles  of  the  chord  mem- 
bers to  the  center  of  gravity  axis,  parallel  to  it,  of  the  rivets. 

My      36,700  x  2.25      10  onn  „ 
For  Web  Members,  /  =  —~  =  -  —  -  12,300  Ibs.  per 

1  D.o 

sq.  in.    The  2.25  ins.  is  the  distance  from  the  back  of  the  angle 
to  the  center  line  of  the  rivets. 

Compare  the  stresses  due  to  the  eccentricity  caused  by  failing 
to  have  the  lines  through  the  center  of  gravity  of  the  members 
meet  at  a  common  po  nt,  with  the  stresses  used  in  the  design  and 
marked  on  Fig.  115.  It  will  be  seen  that  the  secondary  stresses 
are  about  fifty  per  cent  greater  than  the  direct  stresses  and  will 
cause  the  failure  of  the  joint  ultimately,  for  the  effect  of  the  direct 
and  eccentric  stresses  is  the  sum  of  the  two. 


D  must  not  be 


JOINTS   AND   CONNECTIONS  199 

In  proportioning  the  web  member  carrying  tension  the  area 
of  one  rivet  hole  is  deducted  from  the  area  of  the  member.  In 
the  web  member  subjected  to  compression  the  whole  area  is  taken, 
for  the  rivet  is  assumed  to  fill  the  hole.  The  resulting  direct 
stresses  are  shown  in  the  figure.  Provided  the  rivets  are  driven 
on  the  line  of  the  center  of  gravity  of  the  member  the  ascer- 
tained direct  stresses  will  be  realized.  If  the  rivets  are  driven 
to  one  side  of  the  axis  of  the  center  of  gravity  an  eccentricity 
will  be  developed  which  may  very  seriously  increase  the  stress. 
Where  angles  are  used  to  resist  direct 
stress,  and  connected  through  one  leg 
only,  the  gauge  line  for  the  rivets  should 
be  set  in  as  close  to  the  back  of  the 
angle,  or  as  near  to  the  center  of  gravity 
axis  as  possible.  This  matter  is  of  funda- 
mental importance  and  yet  it  is  habitu- 
ally disregarded  in  structural  work. 

The  rivet  clearance  for  machine  driving  is  shown  in  Fig.  116. 
It  is  customary  to  use  so-called  "standard  gauges  "  for  angles, 
pitching  the  rivets  from  the  back  of  the  angle  a  distance  some- 
what greater  than  the  half  width  of  the  leg.  In  the  case  of  the 
web  members  shown  in  Fig.  115  the  dimension  D  would  be  |f 
in.  Adding  to  this  the  thickness  of  the  outstanding  leg,  we  obtain 
1|  ins.  as  the  permissible  gauge  of  these  angles.  This  coincides 
exactly  with  the  center  of  gravity  axis  of  the  angles  and  if  the 
rivets  were  so  placed  the  fiber  stress  due  to  eccentricity  of  the 
line  of  rivets  would  be  entirely  eliminated.  The  method  to  use 
in  figuring  the  stress  due  to  eccentricity  in  the  line  of  rivets  is 
given  in  detail  in  another  chapter. 

To  avoid  eccentricity  as  much  as  possible  in  a  group  of  rivets 
the  forces  should  act  through  the  center  of  gravity  of  the  group. 
The  best  way  to  obtain  the  center  of  gravity  is  by  using  the 
method  of  moments,  as  shown  in  Fig.  117. 

Having  found  the  center  of  gravity  of  a  group  of  rivets  find 
the  axis  through  the  center  of  gravity  of  the  member  connected 
to  the  plate  or  angle  forming  the  connection.  Extend  this  axis 
through  the  group  of  rivets.  If  it  does  not  pass  through  the 
center  of  gravity  of  the  group  it  can  be  replaced  by  a  force  equal 
in  amount  and  parallel,  which  will  pass  through  the  center  of 
gravity.  There  will  be  a  moment  developed  equal  to  the  load 


200 


PRACTICAL  STRUCTURAL  DESIGN 


multiplied   by  the  perpendicular   distance  between  these  two 
lines  of  action. 

Each  rivet  in  the  group  carries  an  amount  of  direct  stress  as 
shear  equal  to  the  total  load  divided  by  the  number  of  rivets. 

STATICAL  MOMENT  ABOUT 
AXIS  xx' 


Rivet 

Area 

Arm 

Moment 

1 

A 

o 

Ao 

2 

A 

0 

Ao 

3 

A 

d 

Ad 

4 

A 

d 

Ad 

5 

A 

e 

Ae 

?    t     3 

\    <? 
x-i-.t~.~ 


STATICAL   MOMENT  ABOUT 
AXIS   yy' 


Rivet 

Area 

Arm 

Moment 

1 

A 

0 

Ao 

2 

A 

c 

Ac 

3 

A 

e 

Ac 

4 

A 

b 

Ab 

5 

A 

a 

Aa 

Total  Area  =  5A 
Total  M  =A(2d  +  e) 


Total  Area  =  5A 
Total  M  =  A( 


5A 


9  = 


5A 


Fig.  117.  —  Method  of  Moments  to  find  Center  of  Gravity  of  Group  of  Rivets. 

If  the  direct  application  of  the  load  causes  a  bending  moment, 
then  each  rivet  has  an  added  stress  due  to  bending  moment. 

The  stress  in  any  rivet  due  to  bending  moment  varies  directly 
as  its  distance  from  the  center  of  gravity  of  the  group  of 
rivets,  and  its  resisting  moment  varies  as  the  square  of  this 
distance. 

Fig.  118  represents  an  angle  connection  for  the  end  of  a  10-in. 
I-beam  weighing  25  Ibs.  per  lineal  foot.  The  reaction  from  the 
load  carried  on  the  beam  is  13,720  Ibs.  delivered  to  the  leg  out- 
standing from  the  beam,  the  other  leg  being  connected  to  the 
web  of  the  beam.  Finding  the  center  of  gravity  of  the  three 
rivets  and  multiplying  we  get  an  eccentric  bending  moment, 

M  =  13,720  X  3.25  =  56,290  in.  Ibs. 


JOINTS   AND   CONNECTIONS 


201 


The  shear  on  each  rivet  due  to  direct  stress  is 

Each  rivet  has  also 
some  stress  due  to 
bending  moment. 

The  necessary  resist- 
ing moment  is  found 
as  follows: 


13,720 


4573  Ibs. 


in  which 

M  =  resisting  moment 
A  =  stress  in  a  rivet 
due  to  bending. 
x,   y   and  z  represent,  g- 

respectively,  the  moment  arm  for  the  rivets  bearing  these  letters. 
Transposing, 


A  = 


56,290 


=  8650  Ibs. 


+  yz+z2       6.513 

The  stress  in  each  rivet  due  to  bending  is  equal  to  this  figure, 

multiplied  by  the  distance  of  the  rivet  from  the  center  of  gravity. 

Stress  on  x  =  8650  x  1.46  =  12,640  Ibs. 

Stress  on  y  =  8650  x  1.46  =  12,640  Ibs. 

Stress  on  z  =  8650  x  1.50  =  12,980  Ibs. 

Adding  the  direct  shearing  stress  to  these  figures,  by  the  paral- 
lelogram of  forces  drawn  in  Fig.  118  the  resultant  stress  on  rivets 
x  and  y  is  15,600  Ibs.,  as  shown. 

The  web  thickness  of  a  10-in.,  25-lb.  I-beam  is  0.31  in.  The 
bearing  area  of  a  f-in.  rivet  is,  therefore,  0.31  x  0.75  =  0.2325 
sq.  in. 

15,600  Ibs.  divided  by  0.2325  =  62,100  Ibs.  per  sq.  in.  bearing 
stress  on  web  of  beam.  This  is  more  than  three  times  the  amount 
shown  as  permissible  in  the  rivet  table,  where  the  allowable  stress 
in  bearing  is  20,000  Ibs.  per  sq.  in.  The  connection  here  shown 
is,  therefore,  not  good  for  a  reaction  of  13,720  Ibs. 

In  Fig.  119  at  (a)  is  shown  an  angle  connected  by  both  legs 
to  the  gusset  plate.  Only  one  hole  is  deducted,  for  the  holes  are 
staggered  to  preserve  the  net  section.  This  staggering  is  done 
also,  when  not  necessary  to  preserve  the  net  section,  in  order  to 
permit  of  driving  the  rivets  in  two  legs  with  plenty  of  "clearance." 


202 


PRACTICAL  STRUCTURAL  DESIGN 


When  an  angle  is  connected  by  both  legs  the  total  area  of  the 
angle,  less  the  area  of  the  holes,  is  taken.  When  an  angle  is 
connected  through  one  leg,  as  at  (6),  the  load  will  be  eccentric. 
The  problem  is  then  complicated,  but  an  easy  practical  solution 
is  to  use  only  the  area  of  one  leg. 

Problem.  An  angle  in  tension  has  to  carry  a  load  of  50,000  Ibs., 
using  f-in.  rivets. 


Area  of  member 


/       16,000 


3.12  sq.  ins. 


W 


If  connected  by 
the  two  legs  try  a 
5"  X  5"  x  |",  the 
area  of  which  is  3.61 
sq.  ins.  (Carnegie, 
p.  146.)  The  area  of 
the  hole  is  f  "  x  f  = 
H"  =  0.328  sq.  in. 
The  net  area  is  3.61 
-  0.328  =  3.282  sq. 
ins.  Notice  that  the  diameter  of  the  hole  is  |"  greater  than  the 
diameter  of  the  rivet,  giving  -fa"  clearance.  This  rule  is  general 
for  all  rivets. 

If  connected  by  one  leg  it  is  assumed  that  this  leg  will  carry 
all  the  stress.  Assume  a  thickness  of  f",  and  as  the  area  is  3.12 
sq.  ins.  divide  and  add,  to  the  width  thus  obtained,  the  diameter 
of  the  hole.  Then  3.12  -=-  0.5  =  6.24"  +  0.875"  =  7.115".  This  is 
plainly  not  suitable,  for  it  does  not  fit  any  standard  angle.  Try 
a  thickness  of  f"  and  we  get  3.12  -=-  0.625  =  5"  +  0.875  =  5.875". 
This  is  nearly  six  niches,  so  we  will  use  a  6"  X  3?"  X  f  "  angle. 

The  practice  of  using  |-in.  and  -fVin-  gusset  plates  in  roof  trusses 
is  very  common,  yet  considerations  of  economy,  as  well  as  effici- 
ency, would  seem  to  indicate  the  use  of  thick  plates.  The  plates 
should  be  of  such  thickness  that  the  bearing  value  of  a  rivet  in 
the  plate  is  about  equal  to  the  value  of  the  rivet  in  double  shear. 
This  would  reduce  the  number  of  rivets  in  a  joint  considerably 
and  reduce  the  size  of  the  plate  correspondingly.  The  slight  in- 
crease in  the  weight  of  the  plates  is  apt  to  be  more  than  offset 
by  the  reduction  in  the  number  of  rivets.  The  use  of  thicker 
places  and  fewer  rivets  also  measurably  reduce  the  secondary 
bending  stresses  in  the  members  due  to  fixity  of  their  ends.  The 


JOINTS  AND   CONNECTIONS 


203 


author  never  uses  a  plate  as  thin  as  \  in.  The  least  thickness  of 
any  member  or  plate  in  a  truss  should  be  T5^  in.,  for  some  metal 
should  be  provided  to  offset  the  wasting  effect  of  rust.  It  is  well 
enough  to  consider  that  the  metal  will  always  be  inspected  and 
kept  painted,  but  we  know  that  painting  is  neglected  for  shame- 
fully long  periods  of  time. 

In  Fig.  120  (a)  is  shown  a  joint  in  a  riveted  Pratt  truss  that  is 
of  common  occurrence.    Here  the  axes  of  the  members  are  con- 


U      (a) 


Fig.  120 


current,  but  the  rivet  connections  through  the  chord  are  eccentric 
to  the  intersection  of  the  lines  of  stress,  and  a  bending  moment 
results.  The  proper  construction  of  this  joint  is  shown  at  (6) 
and  the  student  is  advised  to  perform  the  calculations  for  the 
two  joints  and  determine  for  himself  the  amounts  of  the  eccentric 
stresses.  The  truss  for  which  joints  were  designed  in  wood  can 
be  designed  now  for  steel  and  the  two  details  in  Fig.  120  can  be 
assumed  for  this  truss. 

Fig.  121  (a)  shows  the  heel  of  a  roof  truss.  It  is  a  common 
detail,  but  that  does  not  make  it  desirable  or  proper.  It  merely 
shows  the  power  of  example  and  illustrates  the  proneness  of 
draftsmen  to  copy  blindly.  The  three  forces  acting  at  the  heel, 
namely,  the  compression  in  the  rafter,  the  tension  in  the  bottom 
chord,  and  the  column,  or  wall,  reaction  are  non-concurrent.  A 
bending  moment  results  which  induces  large  fiber  stresses  in  the 
members.  The  method  to  follow  in  determining  the  amounts  of 
the  stresses  was  illustrated  in  the  case  of  the  Warren  truss. 

Fig.  121  (6)  is,  likewise,  an  improper  detail  unless  the  heel 
plate  is  thick  enough  to  resist  the  bending  moment  between  the 
point  of  intersection  of  the  three  forces  and  its  attachment  to 
the  members.  The  plate  should  also  be  planed  or  chipped  flush 
with  the  backs  of  the  angles  of  the  bottom  chord  when  it  is  not 
possible  to  get  sufficient  rivets  immediately  over  the  column  to 
transmit  the  total  reaction  into  the  plate. 


204 


PRACTICAL  STRUCTURAL  DESIGN 


Fig.  121  (c)  show  an  efficient  and  proper  detail  for  the  heel  of 
a  truss. 

Fig.  122  (a)  shows  a  detail  of  a  knee  brace  connection  to  a 
column  which  is  not  uncommon  in  mill  building  construction. 


(c) 


Fig.  121 


Fig.  122 


This  detail  is  open  to  the  same  criticism  as  the  other  eccentric 
connections  already  discussed.  It  is  especially  to  be  condemned 
in  view  of  the  fact  that  the  knee  brace  is  subject  to  tension,  as 
well  as  compression,  and  when  the  knee  brace  is  in  tension  the 
entire  stress  must  be  resisted  by  two  rivet  heads. 

Fig.  122  (6)  shows  the  proper  detail  for  this  connection.  The 
gauge  A,  for  the  rivets  connecting  the  knee  to  the  column  flange, 
should  be  as  small  as  possible,  and  the  thickness  of  the  connection 
angles  should  be  such  that  their  moment  of  resistance  at  the 
rivets  is  equal  to  the  bending  moment.  This  bending  moment 
is  equal  to  one-half  the  horizontal  component  of  the  stress  in  the 
knee  brace,  multiplied  by  A. 

Fig.  122  (c)  is  a  detail  known  as  a  knuckle  plate  connection. 


JOINTS  AND   CONNECTIONS  205 

It  is  sometimes  used,  in  lieu  of  a  knee  brace,  in  order  to  economize 
head  room  and  to  avoid  obstructing  the  crane  trolley  travel.  The 
knuckle  plate  should  never  be  used  as  a  substitute  for  the  knee 
brace  in  a  building  high  enough  for  a  crane. 

The  greater  part  of  the  material  and  all  but  three  of  the  illus- 
trations in  the  section  on  "Secondary  Stresses  in  Framed  Stresses" 
was  taken  from  a  paper  bearing  that  name  presented  by  E.  W. 
Pittman  before  the  Engineers  Society  of  Western  Pennsylvania. 
In  several  paragraphs  the  exact  language  was  used.  The  paper 
was  published  in  the  Proceedings  of  the  Society  (Pittsburgh,  Pa.) 
for  February,  1909.  The  student  should  read  the  whole  of  the 
paper  and  the  discussion  following  it.  In  the  December,  1916, 
issue  of  the  Journal  appeared  a  valuable  paper  by  E.  W.  Pittman 
entitled  "  Factors  Affecting  Cost  of  Structural  Steel  Work,"  and 
a  paper  by  George  H.  Danforth  entitled  "Some  Items  Affecting 
Cost  of  Structural  Steel  Work,"  with  discussion  by  a  number  of 
men  of  wide  experience.  Back  numbers  of  the  Journal  cost  fifty 
cents  each. 

Reference  was  made  in  an  earlier  chapter  to  Smoley's  Parallel 
Tables  of  Logarithms  and  Squares,  which  are  indispensable  to 
structural  designers  in  computing  the  lengths  and  bevels  of  truss 
and  frame  members.  During  the  month  of  January,  1917,  ap- 
peared Smoley's  Parallel  Tables  of  Slopes  and  Risers,  with  Ready 
Solution  of  Right  Triangles.  This  second  book  also  will  be  one 
that  structural  designers  will  not  willingly  go  without,  once  they 
learn  its  value. 

The  design  of  compression  members  will  be  dealt  with  in  the 
chapter  on  Columns  and  Structures.  In  Chapter  VI  will  be 
taken  up  also  the  design  of  members  subjected  to  both  tension 
and  compression. 

The  "One-book  Man"  is  not  a  broad  man.  Even  with  the 
best  possible  explanations  an  author  does  not  always  succeed  in 
getting  the  reader  to  thoroughly  grasp  his  ideas.  It  is,  therefore, 
an  excellent  plan  to  do  some  collateral,  parallel  reading  when 
studying,  in  order  to  obtain  the  methods  of  working  of  more  than 
one  person.  The  author  believes  students  will  receive  a  great  deal 
of  benefit  by  starting  at  this  point  to  study  the  following  books. 

"Bridge  and  Structural  Design."    Thomson,  $2. 

"Structural  Engineering."    Husband  &  Harby,  $2.50. 

"Typical  Steel  Railway  Bridges."    Thomson,  $2. 


206  PRACTICAL  STRUCTURAL  DESIGN 

The  books  named  abound  in  worked  examples  of  detailing. 
The  authors  had  in  mind  men  of  Limited  mathematical  attain- 
ments. 

Not  all  the  students  will  become,  or  wish  to  become,  structural 
detailers.  To  become  a  first-class  commercial  detailer  will  re- 
quire a  great  deal  of  practical  experience  and  the  following  books 
should  be  studied  and  kept  as  works  of  reference. 

"Steel  Structures."    Morris,  $2.25. 

"Structural  Engineers'  Handbook."    Ketchum,  $5. 

The  only  really  adequate  book  on  the  design  of  modern  high 
steel  frame  buildings,  popularly  called  "sky-scrapers,"  is  "Steel 
Construction,"  by  H.  J.  Burt,  $2.25.  It  is  written  in  a  simple 
manner  for  the  instruction  of  correspondence  school  students. 
The  books  by  Morris  and  Ketchum  are  of  college  grade.  The 
publishers  of  this  book  can  supply  all  the  books  mentioned. 


CHAPTER  VI 
Graphic  Statics 

THE  student  knows  that  a  line  can  be  drawn  to  represent  a 
force,  because  forces  act  through  the  center  of  gravity  of 
a  body  and  this  is  a  point.    A  line  is  a  succession  of  points, 
or  is  the  path  of  travel  of  a  point.    A  line  representing  a  force 
indicates  by  its  direction  the  direction  in  which  the  force  acts. 
When  drawn  to  a  scale  the  length  may  represent  the  amount  of 
the  force  in  any  selected  unit,  —  pounds,  tons,  etc. 

In  earlier  chapters  some  simple  graphical  methods  were  pre- 
sented for  obtaining  bending  moments  and  shear  on  beams,  but 
to  solve  more  complicated  problems,  and  to  make  even  those 
shown  a  little  more  A 
simple,  reciprocal  dia- 
grams must  be  used. 
The  reciprocal  dia- 
gram method  for 
making  computations 
is  known  by  the  name 
of  Graphic  Statics. 

In  Fig.  123  is  shown 
the   Parallelogram    of  Fig.  123.  -  Parallelogram  of  Forces. 

Forces.  The  line  AB  shows  the  pressure  of  the  wind  against  a 
roof  truss.  All  forces  act  normally  to  the  surface  pressed,  so  the 
wind  always  blows  perpendicularly  to  the  slope  of  a  roof.  It  may 
be  resolved  into  a  vertical  force  represented  by  BD  and  a  hori- 
zontal force  represented  by  the  line  BC. 

The  vertical  force  as  weight  may  be  added  to  the  weight  of 
the  roof  and  any  loads  that  may  come  upon  it,  and  the  members 
of  the  truss  be  proportioned  to  carry  all  the  loads.  The  horizon- 
tal force  may  be  assumed  to  act  to  push  the  roof  off  the  support 
and  is  a  measure  of  the  force  to  be  resisted  by  bolts  which  tie  the 
truss  to  the  supports. 

All  forces  may  be  resolved  into  components.  The  force  A  B 
in  Fig.  123  might  be  considered  as  produced  by  two  given  forces, 

207 


208 


PRACTICAL  STRUCTURAL  DESIGN 


Fig.  124.  —  Triangle  of  Forces. 


BC  and  BD.  That  is,  if  the  horizontal  force  had  been  given,  the 
line  CB  could  have  been  drawn.  The  vertical  force  having  been 
given,  the  line  BD  could  have  been  drawn,  it  being  assumed  that 
the  two  forces  meet  at  one  point,  B.  From  D  draw  a  horizontal 
line  and  from  C  draw  a  vertical  line  to  intersect  it,  at  A.  The 
diagonal  line  AB  is  the  resultant  of 
the  two  forces. 

The  resultant  is  the  line  required 
to  close  a  figure  and  the  other  forces 
are  components  of  the  resultant. 

In  Fig.  124  it  is  shown  that  it  is 
not  necessary  to  complete  the  parallelogram.  Let  AB  and  A  B' 
be  two  forces  acting  at  the  point  A.  Draw  them  to  scale  and 
from  B'  draw  a  line  parallel  to  AB.  From  B  draw  a  line  parallel 
to  AB'.  The  two  lines  will  meet  at  C.  The  line  AC  is  the  result- 
ant. The  force  A  B'  is  the  force  BC.  The  resultant  would  have 
been  obtained  if  we  had  merely  drawn  AB  and  then  from  B 
drawn  BC  to  represent  the  force  AB'.  The  figure  ABC  is  known 
as  the  Triangle  of  Forces.  It  is  formed  by  drawing  the  forces 
end  to  end  and  obtaining  the  resultant  by  drawing  a  closing  line. 
In  order  to  obtain  equilibrium  it  is  necessary  that  all  the  forces 
acting  on  any  body  meet  at  a  common  point  or  be  parallel.  If 
all  non-parallel  forces  do  not  meet  at  a  point  then  there  will  be  a 

moment  arm  which  will  cause 
rotation.  When  all  the  forces 
are  parallel  equilibrium  is  ob- 
tained by  supplying  a  result- 
ant large  enough  to  balance 
the  forces  acting  in  opposite 
directions. 

In  Fig.  125  are  four  forces 
acting  at  the  point  0,  of  a 
body.  The  polygon  of  forces 
is  shown  at  the  right  and  all 
the  lines  are  drawn  to  scale.  The  line  04  drawn  to  close  the 
polygon  represents  the  weight  which  the  body  must  have  to  resist 
being  moved  by  the  forces.  It  represents  also  the  direction  in 
which  the  body  will  move  if  not  heavy  enough  to  resist  the  push. 
This  polygon  could  have  been  drawn  as  follows,  and  the  student 
is  advised  so  to  draw  it  in  order  to  get  a  clear  idea  of  the  matter ; 


Fig.  125.  —  Polygon  of  Forces. 


GRAPHIC   STATICS  209 

First,  taking  the  forces  1  and  2,  construct  a  parallelogram  and 
obtain  the  resultant.  Using  this  resultant  as  a  force  form  a 
parallelogram  with  it  and  force  3.  The  resultant  thus  obtained 
will  be  combined  as  a  force  with  force  4.  The  final  resultant 
obtained  will  be  the  force  04.  On  the  figure  here  shown  the 
resultant  of  1  and  2  will  be  a  line  02.  The  resultant  of  02  and 
force  3  will  be  a  line  03.  Then  the  resultant  of  force  4  and  03 
will  be  04. 

Action  and  reaction  are  equal  when  equilibrium  is  to  be  pre- 
served. The  arrow  points  on  the  force  lines  show  the  direction 
in  which  each  force  acts.  The  resultant  measures  the  force  neces- 
sary to  preserve  equilibrium;  therefore,  the  direction  is  against 
the  general  direction  of  all  the  forces.  The  arrow  point  on  the 
resultant  indicates  this,  and  the  result  is  that  the  closed  figure 
has  the  arrow  points  so  arranged  that  the  forces  can  be  followed 
consecutively.  This  indicates  equilibrium,  and  if  the  arrow 
points  do  not  indicate  a  consecutive  line  of  travel  it  is  evidence 
that  equilibrium  does  not  exist.  The  polygon  should  not  be 
closed,  or  there  is  some  mistake  in  the  construction. 

The  resultant  is  the  force  required  to  maintain  equilibrium, 
or  it  may  be  a  force  which  can  replace  all  the  other  forces.  In 
Fig.  123  is  an  example  of  where  two  forces  are  substituted  for  one 
force.  The  vertical  and  horizontal  components  may  be  assumed 
to  replace  the  diagonal  force  acting  against  the  roof.  Here  there 
is,  strictly  speaking,  no  resultant  considered.  A  certain  force  has 
been  resolved  into  two  components. 

To  obtain  a  resultant  all  the  forces  are  arranged  to  form  a  tri- 
angle or  polygon  and  the  closing  line  is  the  resultant.  To  resolve 
a  force  into  two  forces  acting  at  any  angle  draw  the  resultant  to 
scale.  From  one  end  draw  a  line  of  indefinite  length  in  the  direc- 
tion of  one  component.  From  the  other  end,  on  the  same  side, 
draw  another  line  of  indefinite  length  in  the  direction  of  the 
second  component.  The  lines  will  intersect  and  the  lengths  thus 
fixed  will  represent  to  scale  the  amounts  of  the  components.  The 
student  is  advised  to  study  carefully  the  difference  between 
resultant  forces  and  component  forces.  A  force  may  have  any 
number  of  components,  but  in  a  system  of  framing  the  number 
will  be  fixed  by  the  number  of  members  meeting  at  a  joint. 

The  designer  of  buildings  will  usually  deal  only  with  parallel 
loads.  The  direct  loads  on  a  roof  act  vertically  at  the  joints.  The 


210 


PRACTICAL  STRUCTURAL  DESIGN 


wind  load  will  also  act  at  the  joints,  but  not  vertically,  this,  how- 
'ever  making  no  difference  in  the  method  of  treatment,  for  the 
wind  loads  on  the  joints  will  be  parallel,  although  not  vertical. 
In  an  earlier  chapter  the  wind  effect  on  roofs  was  assumed  to  be  a 
vertical  load,  but  when  finding  the  stresses  in  a  roof  truss  by 
graphic  statics  one  diagram  is  drawn  for  the  vertical  loads  and 
a  separate  diagram  for  the  effect  of  wind.  Stresses  are  tabulated 
for  each  system  of  loading  and  added.  There  are  several  "com- 
bined "  methods  for  making  a  single  diagram  serve  for  the  vertical 
loads  and  wind  loads,  but  they  are  not  easy  to  remember  and  the 
separate  diagrams  cannot  be  forgotten  once  they  are  mastered. 


Truss 


Fig.  126.  —  Forces  in  King  Truss  by  Graphic  Statics. 

In  Fig.  126  is  shown  a  king  truss  with  the  reciprocal  diagram. 
The  loads  are  marked  at  each  joint.  The  two  end  loads  are  carried 
directly  by  the  walls,  so  do  not  affect  the  stresses  in  the  members 
of  the  truss.  The  spaces  between  members  are  lettered.  A  verti- 
cal line  is  drawn  on  one  side  of  the  truss.  This  is  usually  on  the 
right  side,  but  it  may  be  on  the  left  if  most  convenient.  Begin- 
ning at  the  top,  all  the  loads  are  set  off  as  shown.  The  loads  are 
between  the  lettered  portions  of  the  members,  so  the  spaces  on 
the  vertical  load  line  drawn  to  scale  represent  the  loads  between 
the  letters.  After  setting  off  all  the  loads  the  lines  of  the  truss 
members  are  transferred  by  means  of  triangles,  or  parallel  rule, 
to  connect  with  the  load  line.  This  gives  the  sloping  lines  meet- 
ing on  the  line  CF. 

The  construction  here  described  is  for  a  truss  uniformly  loaded 
with  equal  reactions.  The  amount  of  the  reaction  at  each  end' 


GRAPHIC  STATICS  211 

is  shown  on  the  right  of  the  load  line.  The  loads  are  set  off  verti- 
cally to  scale  and  when  the  construction  of  the  stress  (reciprocal) 
diagram  is  completed  the  lengths  of  the  lines  are  measured  by 
the  same  scale  and  the  amounts  of  the  stresses  found. 

The  character  of  the  stress  in  each  member  must  be  determined 
and  this  can  only  be  done  by  some  study  and  by  following  the 
loads.  It  is  a  help,  at  first,  to  make  a  free-hand  diagram  for  each 
joint  and  place  arrow  heads  on  each  line,  remembering  that  if 
the  frame  is  to  be  in  equilibrium  it  must  be  possible  to  start  at 
one  joint  and  follow  the  lines  clear  around  the  figure  to  the  start- 
ing point.  The  student  is  advised  to  make  the  figures  now  to  be 
described,  free  hand,  and  move  the  pencil  as  he  follows  the 
directions. 

The  kind  of  stress  is  to  be  found.  Start  with  the  joint  between 
B  and  D,  sketching  it  on  the  stress  diagram.  Here  there  is  a 
vertical  load  of  2000  Ibs.,  from  B  to  D  on  the  reciprocal  (stress) 
diagram.  Move  from  D  to  E  and  the  arrow  point  is  seen  to  point 
to  the  joint.  The  stress  is  compressive.  Move  from  E  to  C. 
The  arrow  point  is  towards  the  joint,  so  the  stress  is  compressive. 
Move  from  C  to  B.  The  arrow  point  is  towards  the  joint  and  the 
stress  is  compressive. 

Take  the  joint  at  the  foot  of  the  truss,  between  A  and  B.  This 
should  have  been  taken  first,  but  the  second  joint  gave  the  best 
illustration  for  a  beginner.  Start  from  B  on  the  load  line  and 
move  to  C.  This  is  towards  the  joint,  so  the  stress  is  compressive. 
Move  from  C  to  F  on  the  load  line.  This  is  away  from  the  joint, 
so  the  stress  is  tensile. 

It  has  been  determined  that  the  stress  in  DE,  in  D'E',  in  CE, 
and  in  C'E',  is  compressive.  A  vertical  member  connects  the 
upper  and  lower  joint.  Assuming  this  member  to  be  cut,  will 
the  joints  be  spread  apart  or  will  they  close  up?  If  they  will  be 
spread  apart,  then  the  action  of  the  forces  at  the  joints  will  cause 
tension  in  the  member.  If  they  will  close  up,  then  the  member 
will  be  in  compression.  Applying  this  reasoning,  it  is  seen  that 
the  stress  in  E  E'  is  tensile.  The  character  of  stresses  in  all 
members  of  trusses  may  be  determined  by  such  reasoning. 

The  stress  in  the  vertical  member  may,  however,  be  deter- 
mined by  the  first  method.  On  the  load  line  take  the  load  of  2000 
Ibs.  between  D  and  D'.  Go  from  D'  on  the  load  line  to  E"  and  as 
the  arrow  point  is  towards  the  joint  the  stress  is  compressive. 


212 


PRACTICAL  STRUCTURAL  DESIGN 


From  E'  to  E  the  arrow  point  is  away  from  the  joint,  so  the  stress 
is  tensile.  .  From  E  to  D  the  arrow  point  is  towards  the  joint,  so 
the  stress  is  compressive. 

The  truss  is  symmetrical  with  symmetrical  loads,  so  having 
obtained  the  stresses  and  their  character  in  the  members  on  one 
side  of  the  center,  as  well  as  in  the  center  vertical  member,  all 
the  required  information  has  been  found. 

In  Fig.  127  is  shown  a  Queen  truss  with  its  reciprocal  diagram. 
The  truss  is  symmetrical  with  symmetrical  loading,  so  the  end 
reactions  are  equal.  It  is  only  necessary  to  draw  one  half  of  the 
stress  diagram,  for  there  is  no  center  vertical  member.  ABC  I  A 


k- -9000 

Fig.  127.  —  Forces  in  Queen  Truss  by  Graphic  Statics. 

is  the  reciprocal  for  the  joint  between  A  and  B.  BCEDB  is  the 
reciprocal  for  the  joint  between  B  and  D.  ICE  HI  is  the  recip- 
rocal for  the  joint  CEHL  DFGHED  is  the  reciprocal  for  the 
joint  EGH.  FF'GF  is  the  reciprocal  of  the  joint  between  F  and 
F'.  The  point  on  the  load  line  is  below  the  point  7,  a  distance 
equal  to  FI. 

In  the  examples  shown  no  account  has  been  taken  of  wind. 
The  reactions  are  equal  and  when  wind  is  considered  as  acting 
normal  to  the  surface  of  the  roof  the  reactions  are  unequal. 
Therefore  until  the  graphical  method  for  obtaining  reactions 
with  unequal  loading  is  shown,  we  will  assume  the  wind  to  be 
reduced  to  a  load  acting  vertically.  For  slopes  not  exceeding 
30  degrees  from  the  horizontal  this  will  not  make  much  difference 
and  is  fairly  common  practice.  To  avoid  confusing  the  student 
vertical  loading  will  be  considered  first  and  after  illustrating  the 
work  on  the  forms  of  trusses  in  common  use  the  general  method 
for  obtaining  stresses  caused  by  wind  will  be  taken  up. 

The  word  "  stress "  has  been  used  because  it  is  so  commonly 
used  in  connection  with  the  members  of  a  truss.  It  is  not  strictly 
correct.  "  Force  "  is  the  proper  word  and  forces  act  on  the  mem- 


GRAPHIC   STATICS 


213 


bers.  These  forces  set  up  strains  in  the  members.  A  strain  is  the 
amount  of  deformation  caused  by  the  action  of  an  external 
force  and  all  internal  stresses  are  caused  by  strains.  From  this 
point  the  word  "  force  "  will  be  used  instead  of  the  word  stress." 

The  loads  on  the  panel  joints  are  found  by  assuming  an  area 
equal  to  the  width  of  a  panel  and  a  length  equal  to  the  spacing 
between  trusses.  This  is  based  on  each  truss  carrying  the  load 
on  a  distance  measured  halfway  between  adjacent  trusses  and 
each  panel  in  a  truss  carrying  a  load  on  a  width  measured  half 
way  between  adjacent  verticals.  Half  the  load  on  the  end  panels 
of  a  truss  is  carried  by  the  wall. 

Fig.  128  shows  a  truss  with  a  middle  vertical  and  two  side 
verticals,  together  with  one-half  the  force  diagram.  The  con- 


Fig.  128.  —  Forces  by  Graphic  Statics  in  Queen  Truss  with  Center  Rod. 

struction  is  not  different  from  the  trusses  already  illustrated,  but 
attention  is  called  to  the  fact  that  the  side  verticals,  in  the  recipro- 
cal diagram,  end  at  the  tie  rod.  The  middle  vertical,  on  the  con- 
trary, ties  the  truss  together  —  speaking  in  a  general  way  —  so 
it  is  measured  from  G  to  G'.  The  student  should  be  able  without 
trouble  to  trace  the  reciprocals  on  Fig.  128. 

Fig.  129  illustrates  a  truss  with  a  cambered  tie  rod.  Notice 
that  the  loads  are  plotted  in  order  on  the  reciprocal  diagram  and 
the  sloping  lines  transferred  as  before.  From  the  middle  point 
there  are  two  reaction  lines,  representing  the  camber  in  the  tie 
rod,  instead  of  one  horizontal  line.  The  side  verticals  end  on  the 
lines  representing  the  tie  rod,  but  the  middle  vertical  goes  across 
the  open  space  to  tie  the  two  main  rafters  together  in  the  top 
chord  and  carry  the  stress  to  the  tie  rod.  If  this  vertical  line 


214 


PRACTICAL  STRUCTURAL   DESIGN 


were  not  drawn  in  this  way  there  would  be  a  vertical  line  with  a 
gap  in  it  and  the  letter  m  and  the  letter  I  would  be  repeated. 
Since  there  is  but  one  member  with  these  letters  there  can  be 
but  one  line  in  the  reciprocal  diagram  having  these  letters  at 
the  ends. 

In  Fig.  129  is  used  for  the  first  time  the  most  orderly  arrange- 
ment of  letters  to  designate  truss  members.  Beginning  at  the 
left  the  letters  run  consecutively  along  the  outside  of  the  truss  to 
the  extreme  right  and  continue  in  the  same  consecutive  manner 
back  to  the  left.  When  the  loads  are  laid  off  on  the  vertical  load 
line  the  letters  run  in  regular  order.  This  is  a  convenient  way 


Fig.  129.  —  Truss  with  Cambered  Tie  Rod. 

when  the  truss  is  unsymmetrically  loaded  and  is,  therefore,  con- 
venient when  it  is  symmetrically  loaded  and  but  one-half  of  the 
reciprocal  diagram  is  shown.  Capital  letters  are  used  on  the  truss 
diagram  and  the  corresponding  small  letters  shown  on  the  recip- 
rocal diagram.  The  joints  are  numbered,  in  order  that  the  joint 
referred  to  may  be  described  readily  with  the  fewest  words.  This 
system  for  numbering  joints  is  used  generally  for  trusses  having 
non-parallel  chords.  Trusses  with  parallel  chords  have  the  joint 
numbered  with  the  letter  U  prefixed  to  the  number  on  the 
upper  chord  and  with  the  letter  L  prefixed  to  the  number  on 
the  lower  chord.  The  student  is  advised  to  use  the  method  of 
lettering  and  numbering  shown  in  Fig.  129. 

In  Fig.  130  is  shown  a  simple  Fink  truss  with  cambered  tie  rod. 
The  Fink  truss  is  said  to  have  been  invented  by  Albert  Fink  and  is 
an  early  form  of  metal  truss.  In  Europe  it  is  known  as  the  Belgian 
truss  and  is  commonly  supposed  there  to  have  been  the  invention 
of  a  Belgian  engineer.  In  some  English  text  books  it  is  called  a 


GRAPHIC   STATICS 


215 


truss.  Such  happenings  are  common  in  engineering.  Men  educated 
in  mathematics  and  applied  mechanics,  when  set  to  solve  a  problem, 
are  apt  to  solve  it  in  pretty  much  the  same  way,  although  separated 


Fig.  130.  —  Fink  Truss  with  Cambered  Tie  Rod,  Load  on  Upper  Chord. 

by  oceans  and  continents  and  perhaps  ignorant  each  of  the  other. 
Patent  attorneys  say  that  certain  patents  are  applied  for  at  about 
the  same  time  by  several  men,  the  one  first  making  an  application 
receiving  the  patent,  and  the  others  are  left  with  a  firm  belief  that 
some  one  must  have  told  him  of  their  work. 

The  truss  with  the  cambered  tie  rod  and  diagonal,  instead  of 
vertical,  ties  is  treated  like  the  truss  in  Fig.  129,  but  the  diagonal 
ties  meet  as  shown  at  a  point  on  the  horizontal  center  line. 


Fig.  131.  —  Truss  with  Load  on  Upper  and  Lower  Chords. 

In  Fig.  131  is  shown  a  truss  with  loads  on  the  lower  chord  as 
well  as  on  the  upper  chord.  First  the  loads  on  the  upper  chord 
are  laid  off  on  the  vertical  load  line.  The  loads  on  the  lower 


216  PRACTICAL  STRUCTURAL  DESIGN 

chord  are  added  and  one-half  the  total  gives  the  reaction  for  each 
end.  The  reactions  are  laid  off  as  shown,  overlapping  on  the 
load  line.  The  cambered  tie  rods  are  transferred  to  meet  these 
points  and  extended  to  intersect  with  the  rafter  lines.  The 
horizontal  part  of  the  tie  rod  is  drawn  from  the  middle  point  of 
the  load  line.  The  diagram  is  not  hard  to  construct  if  care  is 
taken. 

The  drawing  for  a  truss  with  a  horizontal  tie  rod,  or  horizontal 
lower  chord,  is  similar.  First  the  loads  on  the  top  chord  are  set 
off  on  the  vertical  load  line.  Then  the  loads  on  the  lower  chord 
are  added,  the  reactions  obtained  and  set  off  and  horizontal  lines 
drawn  from  the  points  marking  the  amount  of  the  end  reactions. 

The  Fink  truss  is  economical  because  the  struts  are  short  and 
most  of  the  members  are  in  tension.  Partial  loading  cannot 
cause  maximum  stresses  in  the  members  as  it  will  in  other  com- 
mon forms  of  trusses.  It  is  a  difficult  roof  to  frame  when  the 
slope  is  slight,  so  it  should  be  used  only  for  pitches  exceeding  27 
degrees.  The  form  shown  in  Figs.  130  and  131  is  the  most  simple 
one,  a  common  form  of  the  Fink  truss  being  illustrated  in  Fig. 
132  and  Fig.  133. 

In  Building  Age  for  May,  1916,  Mr.  Harry  B.  Wrigley,  Allen- 
town,  Pa.,  presented  the  following  method  for  dealing  with  the 
web  members  of  a  Fink  truss,  without  substitution  or  change.  So 
far  as  the  author  knows,  this  method  has  never  before  been  pre- 
sented in  print,  so  Mr.  Wrigley  may  be  justified  in  claiming  it  to 
be  original  with  himself. 

In  the  graphical  analysis  of  forces  in  a  Fink,  or  Belgian,  truss, 
a  difficulty  is  encountered  at  joint  5,  supporting  the  load  CD, 
Fig.  132;  for  after  determining  the  forces  in  the  members  meet- 
ing at  joints  1,  2,  and  3  there  remain  three  unknown  forces  at 
joint  5,  namely,  in  members  DP,  PO,  and  ON.  A  similar  diffi- 
culty is  met  with  at  joint  4,  where  there  are  three  unknown 
forces,  namely,  in  members  NO,  OR,  and  RK. 

It  is  a  well-known  fact  that  in  order  to  construct  a  polygon  of 
forces  in  equilibrium,  acting  in  the  same  plane,  through  the  same 
point,  all  conditions  but  two  must  be  known. 

In  Fig.  132  (a)  is  shown  the  truss,  with  the  force  diagram  at 
(6).  To  illustrate  the  new  method  consider  the  left  half  of  the 
truss  as  shown  at  (c)  and  lay  off  the  load  line  abcdek,  and  reactions 
kr'  and  r'a,  in  the  usual  manner;  then  construct  the  stress  diagram 


GRAPHIC   STATICS 


217 


for  the  web  members  as  shown  by  the  dotted  lines  at  (6),  taking 
the  joints  in  numerical  order. 

It  is  evident  that  the  panel  loads  to  the  right  of  load  EF  do 
not  affect  the  web  members  to  the  left  of  this  load;  therefore, 
having  determined  the  web  forces,  complete  the  load  line  aj, 
and  reactions  jk  and  ka,  at  (6),  to  obtain  the  forces  in  the  chord 
members.  The  diagram  Imnopqr  should  be  made  identical  with 


Fig.  132.  —  The  Wrigley  Solution  of  the  Fink  Truss  Problem. 

the  dotted  diagram  I'm'n'o'p'q'r'.  The  above  method  applies  as 
well  to  diagrams  for  wind  loads. 

In  Fig.  133  three  methods  are  shown  for  drawing  the  reciprocal 
diagram  for  a  Fink  truss.  The  Barr  method  is  to  proceed  with 
the  construction  until  the  line  is  drawn  from  v  to  u.  Then  on  the 
truss  draw  a  dotted  line  from  joint  4  to  joint  6.  On  the  reciprocal 
diagram  from  the  point  u  draw  a  line  towards  t,  of  indefinite  length. 
Transfer  from  the  truss  diagram  the  line  from  4  to  6,  meeting, 
on  the  reciprocal  diagram,  the  line  ui  at  d',  and  terminating  at 
r,  on  the  line  er.  From  r  draw  the  line  rs  and  the  remainder  of 
the  diagram  is  then  readily  drawn.  This  is  applicable  for  cases 
of  unequal  loading  and  for  wind. 

A  method  which  is  correct  when  the  reactions  are  equal  and 
the  joints  are  equally  loaded,  is  to  draw  the  line,  on  the  reciprocal 


218 


PRACTICAL  STRUCTURAL  DESIGN 


diagram,  from  w  to  v  and  then  from  s  to  r.  With  unequal  load- 
ing this  method  must  not  be  attempted.  It  cannot  be  used  for 
wind  loads. 

A  third  method  is  known  as  the  "Moment  solution."  A  truss 
is  merely  a  framed  beam.  Find  the  maximum  bending  moment 
and  divide  by  the  depth  of  the  truss  at  the  point  of  maximum 
moment.  In  the  figure  the  maximum  moment  is  at  the  middle 
of  the  span,  where  the  rise  is  a  maximum.  Dividing  the  moment 
by  the  depth  gives  the  force  in  the  tie  rod  QX.  Set  this  off  to 
scale  from  x  on  the  reciprocal  diagram,  to  q.  Transfer  the  line 
QR  from  the  truss  diagram  to  the  reciprocal  diagram,  from  q 


0>) 

Fig.  133.  — Three  Solutions  of  the  Fink  Truss  Problem. 

to  r.  This  fixes  the  point  r  and  from  here  the  reciprocal  dia- 
gram may  be  completed.  This  method  is  applicable  for  cases  of 
unequal  loading  and  for  wind. 

Unequal  Loading 

In  Fig.  134  is  shown  a  beam  loaded  at  several  points  with  loads 
varying  in  amount.  The  problem  is  to  find  the  reactions.  Draw 
the  beam  to  scale  and  place  the  loads  on  it  at  the  proper  points. 
To  the  right  draw  a  vertical  line  and  set  off  the  loads  to  scale. 

Set  off  a  point,  0,  at  any  distance,  but  the  best  position  is  one 
which  will  make  the  two  lines  drawn  from  it  to  the  end  of  the 
load  line  about  equal  in  length.  This  would  make  the  pole  dis- 
tance, horizontally,  about  half  the  length  of  the  load  line.  The 
scale  will  be  chosen  so  the  diagram  will  not  be  too  large,  provided 
the  scale  used  will  give  all  the  necessary  information.  From  the 
points  on  the  load  line  indicating  the  loads,  draw  lines  to  the  pole. 

Through  the  load  points  on  the  beam  drop  vertical  lines  and 
transfer  the  rays  from  the  polar  diagram  to  form  the  equilibrium 
polygon  at  (6).  The  broken  line  forming  the  bottom  of  the 


GRAPHIC  STATICS 


219 


equilibrium  polygon  must  begin  at  one  reaction  line  and  close  on 
the  other.    The  two  ends  are  connected  by  the  line  oo.    This  line 

is  transferred  to  the  polar  dia- 
gram, as  shown  by  the  dotted 
line.  The  distance  on  the  load 
line  from  the  top  to  an  inter- 
section with  the  closing  line  oo, 
is  the  amount  of  the  left  reac- 
tion and  from  the  point  o,  on 
the  load  line,  to  the  bottom 
of  the  load  line,  is  the  right 
reaction,  all  as  shown  at  (c). 
The  polar  distance,  H,  at 
(c)  is  measured  with  the  scale 
used  in  setting  off  the  loads, 
for  it  is  a  force.  The  vertical 
distances  on  the  equilibrium 


.......  J 

Fig.    134.  —  Moment,   Shear   and    Re- 
actions for  Beam.  ,          ,,        ,  ,. 

polygon,  from  the  closing  line 

to  the  broken  bottom  line,  are  measured  with  the  scale  used  in 
drawing  the  beam.  The  vertical  depth  of  the  equilibrium  poly- 
gon, at  any  point,  multiplied  by  the  polar  distance  gives  the 
bending  moment  at  that  point.  If  the  vertical  distance  is  in  feet 
and  the  polar  distance  is  in  pounds,  the  moment  is  in  pound  feet. 
At  (d)  is  shown  the  shear  diagram,  for  which  it  is  believed  no 
explanation  is  necessary. 


Fig.  135.  —  Roof  Truss  Unequally  Loaded. 

In  Fig.  135  is  shown  a  roof  truss  having  unequal  loads  on  the 
panel  points.    A  polar  diagram  is  drawn  first,  to  obtain  the  reac- 


220 


PRACTICAL   STRUCTURAL  DESIGN 


tions.  The  load  line  is  then  set  off  for  the  reciprocal  diagram 
and  the  reactions  are  scaled  off  on  it.  In  this  case  the  load  line 
must  show  all  the  loads  and  must  be  complete  whereas  in 
trusses  symmetrically  loaded  it  is  necessary  to  draw  only  one- 
half  the  reciprocal  diagram. 
In  Fig.  136  the  treatment  of  a  Warren  truss  loaded  on  the  top 


(V     V 


Fig.  136.  —  Warren  Deck  Truss. 

chord  is  given.  The  vertical  dotted  lines  at  the  ends  are  in 
compression  and  deliver  then*  loads  direct  to  the  abutments. 
The  horizontal  line  to  the  first  joint  in  the  upper  chord  delivers 
part  of  its  load  as  a  reaction,  to  the  upper  joint  and  part  to  the 
vertical  post. 

Notice  carefully  the  treatment  of  the  Warren  through  truss 
(loaded  on  lower  chord),  as  shown  in  Fig.  137. 

In  Fig.  138  is  shown  the  reciprocal  diagram  for  a  Howe  truss 
loaded  on  the  lower  chord  and  in  Fig.  139  is  shown  a  Howe  truss 


CP/   \N/  \L/   \J/   \6 
\    \/0\/M\/K\/l 


V 


\ 


Fig.  137.  —  Warren  Through  Truss. 

loaded  on  the  upper  chord.  In  the  through  truss  the  middle 
vertical  carries  a  load,  but  in  the  deck  truss  it  carries  no  load,  as 
shown  by  the  two  letters  at  the  end  of  the  reciprocal  diagram. 
All  reciprocal  diagrams  must  close  and  when  a  line  must  be 
omitted  in  order  to  make  a  diagram  close,  put  the  extra  letters 
at  the  joint  where  there  is  an  apparent  jumble  and  letter  the  next 


GRAPHIC   STATICS 


221 


joint  in  order.    Reciprocal  diagrams  show  plainly  when  there  is 
a  redundancy  of  members  and  also  show  when  a  member  should 


TH 


Fig.  138.  —  Howe  Through  Truss. 

be  added.  The  student  is  advised  to  make  reciprocal  diagrams 
for  odd  panel  trusses  of  the  Warren,  Howe  and  Pratt  types  after 
studying  this  chapter,  as  an  exercise.  Odd  panel  trusses  have 
been  purposely  omitted  in  order  to  give  the  student  an  oppor- 


Fig.  139.  —  Howe  Deck  Truss. 

tunity  to  exercise  his  brain  in  studying  the  problem.  It  will 
help  to  make  the  complete  diagram  and  run  the  lines  from  the 
two  ends,  so  if  there  is  any  trouble  encountered  it  will  be  caught 
in  the  middle  and  can  be  readily  solved.  In  some  trusses  there 


\ 


Fig.  140.  —  Pratt  Through  Truss. 

may  be  members  that  are  not  stressed  except  under  moving 
loads.     Determine  this  positively.     Take  nothing  for  granted. 


222 


PRACTICAL   STRUCTURAL  DESIGN 


When  reciprocal  figures  do  not  close  there  is  either  another  mem- 
ber required  or  the  work  of  the  draftsman  is  poor. 


\ 


Fig.  141.  — Pratt  Deck  Truss. 

The  diagrams  for  the  Pratt  truss,  shown  in  Fig.  140  and  Fig. 
141  are  readily  followed. 

In  Fig.  142  and  Fig.  143,  of  bowstring  trusses,  the  student 
should  observe  how  nearly  uniform  the  stress  is  in  the  chord  for 
each  panel.  The  web  stresses  are  very  small,  and,  as  the  arch 


Fig.  142.  —  Through  Bowstring  Truss  and  Girder  Stringer. 
more  nearly  approaches  a  parabola  the  lower  will  be  the  stresses 
in  the  web  members  and  the  more  nearly  equal  will  be  the  stress 
in  each  panel  length  of  the  chords.  The  student  should  make 
diagrams  for  Pratt,  Howe,  Whipple  and  Warren  trusses  with 
parallel  chords  and  compare  these  with  the  same  framing  of  web 


Fig.  143.  —  Deck  Bowstring  Truss. 

members  in  bowstring  trusses  of  the  same  span,  carrying  the 
same  loads. 


GRAPHIC   STATICS 


223 


A  form  of  truss  used  for  exhibition  halls,  drill  halls,  etc.,  is 
shown  in  Fig.  144  with  Whipple  framing  of  web  members,  and 
in  Fig.  145,  with  the  web  members  framed  as  in  a  Warren  truss. 


Fig.  144.—  Crescent  Roof  Truss.     Whipple  Framing. 

On  the  load  line  the  loads  are  set  off  vertically  to  scale  and  from 
each  point  a  line  is  drawn  parallel  with  the  top  chord.  The  loads 
being  symmetrical  it  is  really  necessary  to  draw  but  one-half  of 
the  force  diagram.  With  trusses  having  as  complicated  a  fram- 
ing as  these  curved,  crescent  shape  trusses,  it  is  difficult  to  trans- 
fer all  the  lines  from  the  truss  diagram  to  the  force  diagram  and 
have  them  truly  parallel.  In  such  cases  it  sometimes  pays  to 


Fig.  145.  — Crescent  Roof  Truss.    Warren  Framing. 

compute  the  angles  of  slope  by  trigonometry  and  set  the  lines 
off  with  a  protractor  or  by  using  a  table  of  chords. 

The  hog-back  truss  shown  in  Fig.  146  is  in  common  use  and 
sometimes  the  upper  chord  is  curved  instead  of  straight.  No 
difficulty  should  be  encountered  in  obtaining  the  stresses  in  such 


224 


PRACTICAL  STRUCTURAL  DESIGN 


a  truss,  for  it  is  merely  necessary  to  classify  the  web  framing  and 
follow  the  methods  given  for  that  special  framing. 

The  shed-roof  truss  (Fig.  147)  is  so  called  because  it  slopes  one 
way,  like  the  roof  of  a  lean-to  shed.     In  this  type  it  is  necessary  to 

set  off  the  whole 
load  line,  as  though 
the  roof  had  unequal 
reactions.  Notice 
the  difference  in 
the  stresses  in  the 
member  QR  and 
the  member  OP. 

The  scissors  truss 
shown  in  Fig.  148 

is  a  very  common 
Fig.  146. -Hog  Back  Truss.  ^^   ft    fayorite 

with  many  builders  and  draftsmen.  Setting  off  the  loads  on 
the  load  line  it  is  very  quickly  discovered  that  to  make  the 
force  diagram  close  it  is  necessary  to  commence  with  the  load 
on  joint  3,  instead  of  either  joint  2  or  joint  5.  Completing 
the  diagram,  which  has  to  be  done  by  drawing  the  dotted  lines 
aj  and  fj,  it  is  discovered  that  all  the  members  are  in  compression. 
The  dotted  lines  aj  and  fj  represent  the  thrust  of  the  truss  against 
the  walls  or  tops  of  the  buttresses.  The  dotted  line  oj  repre- 
sents the  tension  required  to  resist  the  thrust,  consequently  the 


H  IlL 


Fig.  147.— Shed  Roof  Truss. 

pull  in  a  rod  which  can  be  run  from  joint  1  to  joint  6  and  convert 
the  diagonal  thrust  into  a  vertical  reaction. 

A  vertical  rod  may  be  used  to  connect  joints  3  and  4,  and  this 
will  render  the  rod  from  1  to  6  unnecessary  and  will  convert  the 


GRAPHIC   STATICS 


225 


truss  into  one  having  vertical  reactions.    This  rod  will  also  change 
the  diagram  and  render  it  possible  to  start  from  either  end  of  the 

load  line  and  project  the 
members.  This  the  student 
is  advised  to  do  as  an  exer- 
cise. If  the  reactions  from 
roof  trusses  are  not  vertical, 
walls  will  be  forced  out  and 
the  trusses  will  sag. 

A  step  forward  from  the 
scissors  truss  gives  the  truss 
with  curved  tie,  shown  in 
Fig.  149.    This  is  very  often 
seen    in    churches.      Some- 
times the  framing  is  exposed  and  at 
other  times  a  ceiling  is  attached  to  the 
Fig.  148.  — Scissors  Truss. 

curved  tie. 

The  curved  tie  is  usually 
a  T  iron  which  is  bolted  to 
the  rafters  at  the  ends.  The 
upright  leg  of  the  T  is  set 
into  the  horizontal  brace 
and  into  the  rafters  as  well. 
Though  the  tie  is  curved  the 
pull  is  straight  from  joint  4 
to  joint  3  and  joint  7  and  is 
straight  from  3  to  1  and 
from  7  to  8.  At  (6)  is  shown 
the  truss  diagram  and  at  (c) 
is  shown  the  force  diagram. 
Between  the  joints  the  T  A 
must  have  enough  stiffness  to 
retain  the  curved  form  into 
which  it  is  rolled,  or  bent. 

The  hammer-beam  truss, 
shown  in  Fig.  150,  is  a  hand- 
some truss  much  used  for 
churches  and  stately  halls, 
with  exposed  rafters.  The 


Fig.  149.  — Truss  with  Curved  Tie. 


226 


PRACTICAL  STRUCTURAL   DESIGN 


curved  members  SW  and  NW  have  no  stress  except  under  wind. 
Another  form  of  this  roof  dispenses  with  the  vertical  tie  PQ  and 
the  analysis  resembles  that  for  the  scissors  truss.  Nothing  is 
gained  by  doing  without  the  vertical  tie,  but,  on  the  contrary, 
the  roof  is  bettered  by  having  the  tie. 

In  this  truss  all  that  portion  above  joints  5  and  10  is  treated  as 
a  separate  roof,  resting  on  the  frame  represented  by  joints  1,  5, 
10,  14.  The  dotted  lines  from  5  to  1  and  from  10  to  14  represent 
in  direction  the  thrust  of  the  roof  and  it  is  necessary  to  have 


Fig.  150.  — Hammer  Beam  Truss. 

buttresses  to  carry  it,  or  divide  it  at  the  lower  joints  into  a  ver- 
tical component  which  the  wall  will  carry  and  a  diagonal  com- 
ponent acting  along  the  roof  of  an  aisle  in  the  building.  The 
aisle  roof  in  turn  will  deliver  its  load  to  walls  Or  buttresses. 

In  the  force  diagram,  to  the  right,  the  loads  are  laid  off  ver- 
tically, beginning  with  joint  4,  represented  by  the  load  be,  the  loads 
on  joint  2  and  joint  13  going  vertically  into  the  wall  or  column. 
One-half  the  load  at  joint  4  is  carried  by  the  lower  joint,  so  on 
the  load  line  from  a  point  midway  between  6  and  c  draw  a  dotted 
line  parallel  with  the  line  on  the  truss  diagram  from  5  to  1.  This 
intersects  the  horizontal  reaction  line  at  w,  and  the  length  xw 
gives  the  amount  of  horizontal  thrust  on  the  support  at  5th 
lowest  joint. 

There  is  no  stress  in  rw  and  ow.  These  members  are  stressed 
in  tension  by  the  small  upper  truss  of  which  they  are  the  tie, 
but  they  are  stressed  an  equal  amount  in  compression,  due  to 
the  thrust  against  the  walls,  and  one  stress  neutralizes  the  other. 


GRAPHIC  STATICS  227 

Wind  Force 

It  was  stated  that  when  the  slope  of  a  roof  is  less  than  30 
degrees  it  is  customary  to  assume  the  wind  load  as  acting  hori- 
zontally. When  the  slope  is  greater  than  30  degrees  the  wind 
is  an  important  matter  and  the  exact  amount  and  direction  must 
be  considered.  When  the  forces  in  a  roof  are  treated  graphically 
the  best  practice  is  to  obtain  the  exact  forces  caused  by  wind,  no 
matter  what  the  slope  of  the  roof. 

A  number  of  formulas  for  wind  are  in  use,  but  the  most  modern 
is  that  of  Duchemin.  It  is  based  on  very  careful  experiments 
and  is  considered  the  most  reliable  wind  pressure  formula  now 
in  use. 

Let  P  =  horizontal  wind  pressure  in  pounds  per  square  foot. 
Pn  =  wind  pressure  normal  to  the  surface  of  the  roof,  in 

pounds  per  square  foot. 
A  =  angle  of  the  surface  of  the  roof,  with  the  horizontal, 

expressed  in  degrees. 
»»,    2  sin  A 


All  designers  like  simple  straight-line  formulas,  so  the  following 
is  used  by  a  number  of  men.  It  gives  values  somewhat  lower 
than  those  given  by  the  Duchemin  formula,  but  agrees  fairly 
well  with  some  experiments.  For  roofs  having  a  slope  exceeding 
45  degrees  the  full  horizontal  pressure  is  used.  When  the  angle 
is  less  than  45  degrees,  the  straight-line  formula  is 

Pn  =  P(A  +  45). 

A  number  of  years  ago  Professor  Karl  Pearson  proposed  that  the 
pressure  on  a  roof,  normal  to  the  surface,  be  taken  as  equal  in 
pounds  per  square  foot  to  the  slope  of  the  roof  expressed  in  degrees, 
up  to  a  maximum  of  the  number  of  pounds  horizontal  pressure, 
after  which  the  normal  pressure  should  be  equal  to  the  horizontal 
pressure.  For  example,  when  the  angle  of  the  roof  with  the 
horizontal  is  20  degrees,  the  normal  pressure  will  be  20  Ibs.  per 
square  foot.  Expressed  as  a  formula,  using  a  maximum  of  50 
Ibs.  per  square  foot,  the  horizontal  pressure  used  in  early  days, 
it  appears 

P  *A 


228  PRACTICAL  STRUCTURAL  DESIGN 

and  the  maximum  angle  of  slope  will  be  50  degrees.  Professor 
Ricker  a  few  years  ago  proposed  a  similar  formula,  using  30  Ibs. 
pressure  and  30  degrees  maximum  slope,  to  accord  with  modern 
practice. 

The  horizontal  wind  pressure  is  fixed  in  specifications.  It  is 
usually  taken  as  30  Ibs.  per  square  foot  against  vertical  flat  sur- 
faces. The  following  modifications  are  made  for  surfaces  not 
flat: 

Cylindrical  chimney,  67  per  cent  of  horizontal  pressure. 

Octagonal  chimney,  71  per  cent. 

Rectangular  building  of  large  size,  80  per  cent. 

Concave  side  of  shallow  cylinders,  channels  and  cups  115  to 
130  per  cent.  For  deep  cups  and  concave  side  of  spheres, 
130  to  170  per  cent.  (Ketchum.) 

When  the  wind  pressure  against  a  roof  is  reduced  to  the  normal 
pressure  a  stress  diagram  may  be  drawn  after  finding  the  reac- 
tions. The  pressure  normal  to  the  surface  acts  at  each  joint  the 
same  as  any  load  for  which  the  roof  may  be  designed.  The  effect 
the  wind  force  on  the  roof  will  have  on  the  walls  or  columns 
determines  the  stresses  in  the  roof  members. 

There  are  three  general  cases: 

1.  The  roof  may  be  fastened  to  the  support  at  both  ends. 

2.  The  roof  may  be  attached  to  one  end  support  and  the  other 

end  may  rest  on  a  plate  and  be  free  to  move. 

3.  The  roof  may  be  attached  to  one  end  support  and  the  other 

end  may  rest  on  rollers. 
With  Case  2  and  Case  3 

(a)  The  wind  may  come  from  the  attached  end. 

(b)  The  wind  may  come  from  the  free  end. 

With  Case  1  the  reactions  cannot  be  vertical  and  the  horizontal 
thrust  causes  bending  in  the  roof  support,  whether  it  be  a  wall 
or  a  column.  The  reactions  are  parallel  to  the  resultant  wind 
pressure. 

With  Case  2  the  reaction  at  the  fast  end  is  parallel  to  the  re- 
sultant wind  pressure  and  the  reaction  at  the  "free  "  end  makes 
an  angle  with  the  vertical  equal  to  the  coefficient  of  friction 
between  steel  and  steel,  about  18  degrees. 

With  case  3  the  reaction  at  the  free  end  is  vertical. 

The  cases  selected  for  illustration  are  very  simple.  The  matter 
is  simple.  To  give  a  number  of  force  diagrams  showing  the  effect 


GRAPHIC   STATICS  229 

of  wind  on  a  number  of  forms  of  roof  trusses  would  make  it  appear 
complicated.  A  simple  truss  is  sufficient.  The  principles  are 
as  easy  to  grasp  as  any  of  the  work  in  graphic  statics  and  the 
earnest  student  can  go  through  all  the  trusses  illustrated  in  this 
chapter  and  make  diagrams  for  the  effect  of  wind  on  each  truss. 

First  obtain  the  direction  of  the  resultant  of  the  wind  and  the 
directions  of  the  reactions  due  to  wind.  Draw  a  triangle  repre- 
senting this.  From  the  intersection  where  the  reactions  meet, 
draw  a  reaction  line,  representing  the  forces  in  amount  and 
direction  on  the  lower  chord.  On  the  inclined  wind  resultant 
set  off  the  wind  load  on  each  joint  and  from  this  load  line  draw 
lines  parallel  to  the  members  of  the  truss  and  complete  the  force 
diagram.  The  force  diagram  for  wind  differs  from  that  in  which 
vertical  loads  are  considered,  merely  by  having  the  load  line 
inclined  and  not  vertical.  All  the  other  lines  are  parallel  to  the 
truss  members. 

Tables  of  stresses  must  be  made  for  roof  trusses  when  all  loads 
are  separately  considered.  Such  a  table  will  have  a  number  of 
columns  ruled  on  lined  paper.  Each  system  of  loading  will  have 
two  columns,  one  for  tension  and  one  for  compression.  The 
columns  are  as  follows,  from  left  to  right: 

1.  Designation  of  members  between  joints. 

2.  Dead  load  on  top  chord  (+)  and  (-). 

3.  Snow  load  (+)  and  (-). 

4.  Wind  load  (+)  and  (-). 

5.  Uniform  load  on  lower  chord  (+)  and  (— ). 

6.  Trolley,  or  other,  moving  loads  (+)  and  (-). 

7.  Total  loading  (+)  and  (-). 

Each  member  shown  in  the  last  column  to  have  both  tensile 
and  compressive  forces  to  resist  is  designed  accordingly. 

In  Fig.  151  at  (a)  is  shown  a  roof  truss  and  the  graphical 
method  for  finding  the  reactions  and  the  wind  stresses,  with  the 
two  ends  of  the  truss  secured  to  the  supports.  First,  referring  to 
(a),  the  reactions  are  found  by  multiplying  the  length  of  the  slope 
on  one  side  by  the  distance  between  trusses,  to  obtain  the  area 
acted  upon  by  the  wind.  This  area  is  multiplied  by  the  wind 
pressure  per  square  foot.  It  acts  at  the  center  of  area,  as  shown 
by  the  arrow. 

To  obtain  the  reactions  graphically,  prolong  the  line  of  the 
wind  resultant  through  the  truss,  and  the  length,  ab,  represents 


230 


PRACTICAL  STRUCTURAL  DESIGN 


to  scale  the  amount  of  the  wind  force.  At  the  left  end,  Ri,  lay 
off  the  length  cd,  equal  to  06.  Connect  the  right  end,  Rz,  to  d 
with  a  straight  line,  intersecting  ab  in  e.  Then  the  length  ae 


Fig.  151.  — Wind  on  Roof  with  Ends  Fast. 

represents  the  left  reaction  and  the  length  be  represents  the 
right  reaction.  The  wind  is  assumed  to  be  blowing  from  the 
left,  but,  for  this  case,  the  members  are  dimensioned  to  be  strong 
enough  to  resist  the  wind  from  either  side.  The  reactions  will 
then  in  amount  be  equal  to  the  larger  reaction.  At  (6)  is  shown 
the  force  diagram  for  the  wind. 

In  Fig.  152  is  shown  a  method  for  computing  the  reactions 
when  both  ends  of  the  roof  are  fast  (secured).  The  reaction 
lines  are  drawn  parallel  to  the  wind  resultant.  The  wind  acts 

at  the  center  of  one 
side  of  the  roof.  The 
distances  x  and  y  are 
measured  normal  to 
the  resultant.  Mul- 
tiply the  wind  by  the 
length  y  and  divide  by 


\ 


\ 


f  +  "  the   length   x.      This 

.-""'  gives    R2,    which    is 

subtracted    from    the 
total    wind    force    to 
Fig.  152.— Inclined  Reactions  from  Wind.         obtain  Rr.    The  prob- 
lem is  seen  to  be  that  of  a  beam  carrying  a  single  concentrated 
load. 


Y 


GRAPHIC   STATICS 


231 


In  Fig.  153  the  roof  is  assumed  to  rest  at  one  end  on  rollers, 
in  order  to  take  care  of  temperature  changes,  which,  in  trusses 
secured  at  both  ends,  often  cause  tremendous  changes  in  the 
stresses.  The  re- 
action at  the  free 
end  is  vertical  and 
the  wind  is  from  the 
fast  end.  At  the 

,  ,    , 

free  end  drop  a  ver- 
tical  line.  Through 
the  center  of  area 
on  the  windward 
side  of  the  roof 
draw  a  line,  normal 
to  the  slope,  down- 
ward to  an  inter- 
section with  the 
vertical  reaction 
line.  The  point  of 
intersection  is  then 
connected  to  the 
fast  end  by  a  line, 
which  gives  the  di- 
rection of  result- 


Fig. 153.  —  Wind  Pressure  on  Roof-wind  on  Fast 
Side. 


ant  Ri. 

At  (6)  is  the  force  diagram.  First  draw  a  load  line  parallel 
to  the  wind  resultant  and  lay  off  the  amount  of  wind  at  each 
end  joint  and  at  each  joint  on  the  truss.  From  the  ends  draw 
lines  parallel  to  the  two  reaction  lines.  This  forms  a  triangle 
alf,  the  side  fl  being  equal  to  Rz,  and  the  side  al  being  equal  to 
72 1.  The  remainder  of  the  diagram  is  readily  drawn,  all  the  lines 
on  the  windward  side  being  parallel  to  the  members  of  the  truss, 
with  the  load  line  (the  wind)  inclined. 

In  Fig.  154  is  shown  the  method  to  use  when  the  wind  is  blow- 
ing from  the  free  end  towards  the  fast  end.  No  explanation  is 
required  for  this  figure  if  the  explanations  given  for  Fig.  153  are 
understood. 

The  free  end  of  a  truss  may  rest  on  steel  plates  instead  of 
rollers.  The  only  difference  between  this  method  and  that  when 
the  free  end  rests  on  rollers  is  that  the  reaction  under  the  free 


232 


PRACTICAL  STRUCTURAL   DESIGN 


end  is  inclined  at  an  angle  of  18  degrees  from  the  vertical  away 
from  the  wind,  this  being  practically  the  angle  of  friction  of  steel 
on  steel.  Trusses  are  of  course  designed  for  the  maximum  stresses, 
and  with  the  majority  of  trusses  the  maximum  stresses  occur 

with  the  wind  from 
the  fast  side.  Ana- 
lyze the  truss  with 
wind  from  either  side 
and  then  proportion 
each  member  for  the 
greatest  force  it  is 
send  expected  to  resist, 
niters  the  two  sides  of  the 
truss  being  alike. 


R2 


(b) 


Wind  Pressure  on  Roof -Wind  on  Free 
Side. 


Concentrated  Loads 
Sometimes  a  roof 
truss  must  be  de- 
signed to  carry  a 
trolley  at  some  joint. 
The  designer  does 
not  always  know  in 
advance  on  which 


Fig.  154. 

Sidfi. 

panel  the  trolley  will 

be  carried,  the  owner  of  the  building  wishing  to  be  free  to  change 
such  things  at  pleasure.  Instead  of  a  trolley  it  may  be  a  shaft 
for  machinery,  or  a  heavy  pipe. 

The  method  to  pursue  in  such  a  case  is  to  design  the  truss  for 
the  dead  load,  which  will  include  the  allowance  for  snow  if  any, 
then  design  for  wind,  then  make  diagrams  for  the  concentrated 
loading  at  each  joint  where  it  is  liable  to  come.  This  brings  up 
the  question  of  maximum  and  minimum  stress  and  reversal  of 
stress. 

Maximum  and  Minimum  and  Reversed  Stresses 

Specifications  usually  state  the  safe  allowable  unit  stress  for 
all  materials,  but  seldom  give  the  stresses  to  use  for  members 
subjected  to  changing  stresses  and  reversal  of  stress.  It  is  cus- 
tomary in  such  cases  to  use  a  "  Range  Formula." 


GRAPHIC   STATICS 


233 


Let  /  =  unit  stress  specified,  and  which  will  be  used  for  dead 

load,  or  for  total  quiescent  load. 
Sm  =  maximum  load  on  the  member. 
Sp  =  minimum  load  on  member, 
then 

Working  Stress  =  =TT 


When  one  load  is  compressive  and  the  other  is  tensile  replace 
the  positive  (+)  sign  by  a  negative  (-)  sign. 

Example.  The  maximum  and  minimum  loads  in  a  certain 
member  are  respectively  105,000  Ibs.  tensile  and  49,000  Ibs.  ten- 
sile. What  is  a  proper  working  stress? 


An,    Working  **.  - 


=  13,200  Ibs. 


per  sq.  in. 

Example.  The  maximum  and  minimum  loads  are  respectively 
105,000  Ibs.  tension  and  23,000  Ibs.  compression.  What  is  the 
proper  unit  working  stress? 

Ans.     Working  stress  =  ^^  (l  -  .  N?1>??Ln')   =  9600  Ibs. 
l.O      \         A  X  lUo,UUU/ 

per  sq.  in. 

The  reduced  stress  found  by  the  above  range  formula  is  used 
for  members  in  tension.  The  compressive  stress  is  determined 
by  an  appropriate  column 
formula,  but  it  cannot 
exceed  the  range  stress. 


Snow  Load 

The  snow  load  is  al-   M 
ways  included  with  the   ^ 
dead  load.   It  varies  with 
the  latitude  as  well  as 
with    the    slope    of    the 
roof.       In    Fig.    155    is 
shown  the  snow  load  to 


Latitude  in  Degrees 


Fig.  155.  —  Snow  Load  on  Roofs  for  Different 
Latitudes 


use  according  to  the  recommendation  of  Professor  Ketchum  in 
"The  Design  of  Steel  Mill  Buildings"  in  1903. 

English  text  books  state  that  an  allowance  of  5  Ibs.  per  sq.  ft. 
of  horizontal  projection  is  common  for  Great  Britain.  Ketchum 
recommends  the  minimum  ice  and  sleet  load  for  all  slopes  of 


234 


PRACTICAL  STRUCTURAL   DESIGN 


roofs,  plus  the  recommended  snow  load;  for  a  high  wind  may 
succeed  a  heavy  sleet.  Not  all  engineers  use  the  snow  load  in 
addition  to  the  wind  load,  arguing  that  a  high  wind  will  blow 
away  the  snow.  The  possibility,  however,  of  a  high  wind  follow- 
ing sleet,  which  cannot  be  blown  away,  must  be  considered.  A 
minimum  of  10  Ibs.  per  sq.  ft.  should  be  used  except  in  localities 
mentioned  on  Fig.  155.  A  sleet  storm  may  follow  a  heavy  snow- 
storm, and,  hi  its  turn,  be  followed  by  a  heavy  wind. 


Wind  on  a  Curved  Roof 

In  Fig.  156  is  shown  the  graphical  method  to  follow  in  obtain- 
ing the  reactions  for  a  roof  having  curved  chords.  First  find 

the  inclination  of  each  panel 
of  the  top  chord  in  degrees 
and  find  the  normal  compo- 
nent of  the  wind  on  each 
slope.  Multiply  the  area  of 
each  panel  by  the  normal 
force  of  the  wind  on  the  panel 
and  set  this  off  at  each  end  of 
the  panel  (i.e.,  at  each  joint) 
and  complete  each  parallelo- 
gram of  forces.  Draw  the  dot- 
ted lines,  representing  the  re- 
sultant for  each  parallelogram. 
At  (6)  draw  the  polar  diagram.  The  line  ed  is  parallel  with  the 
resultant  at  the  joint  DEP.  The  line  dc  is  parallel  with  the  re- 
sultant at  the  joint  CDR,  etc.  The  length  of  each  line  is  equal 
to  the  amount  of  the  resultant  wind  force  at  the  joint  through 
which  the  resultant  passes.  Connect  the  points  ae  and  the  direc- 
tion of  the  resultant  wind  force  on  one  side  of  the  roof  is  found, 
and  its  amount. 

At  (c)  draw  the  equilibrium  polygon.  The  line  o2  is  parallel 
with  the  line  ob  of  the  polar  diagram.  Similarly  the  line  23  is 
parallel  with  the  line  oc  and  the  line  3u  is  parallel  with  the  line 
od,  of  the  polar  diagram.  The  line  ou  is  transferred  to  the  polar 
diagram  as  the  resultant  closing  the  equilibrium  polygon.  The 
line  oe  on  the  polar  diagram  is  transferred  to  the  equilibrium 
polygon  as  the  line  w^.and  the  line  oa  is  transferred  as  the  line 


(b) 


Fig.  156 — Reactions  for  Wind  on 
Curved  Roof 


GRAPHIC   STATICS 


235 


0.4.      The  intersection  of  these  lines  fixes  the  location  of  the  re- 
sultant wind  force,  which  acts  normally  to  the  lie  ou. 

The  roof  is  assumed  to  be  fast  at  the  supports,  so  the  resultant 
wind  force  multiplied  by  y  and  divided  by  x  gives  the  amount 
of  #2.  If  one  end  of  the  roof  is  free  the  reactions  are  found  as  in 
Fig.  153  and  Fig.  154,  after  finding  the  amount  and  location 
of  the  resultant  as  just  shown. 

The  remainder  of  the  process  for  ascertaining  the  forces  in 
the  members  is  exactly  as  shown  in  other  cases,  the  only  difference 
being  that  the  load  line  is 
parallel  to  the  resultant  of 
the  wind  pressure.  The  dia- 
gram will  appear  to  be  com- 
plicated but  it  only  needs  care 
and  patience  to  make  it  right. 
Some  of  the  lines  on  the  truss 
diagram  are  very  short  and  it 
may  be  advisable  to  plot  them  A ' 
with  a  protractor,  or  compute 
their  direction  by  using  tables, 
as  it  is  difficult  to  transfer  a 
short  line  and  draw  a  long 
line  parallel  with  it. 

Cantilever  Trusses 
In  Fig  157  is  shown  a  can- 
tilever truss  that  appears  to 
be  a  favorite  with  examiners, 
for  it  is  found  in  many  exami- 
nation papers.  In  this  par- 
ticular example  the  right  re- 
action is  zero.  Sometimes  this 


Fig.  157—  CantUever  Truss 


roof  is  shown  with  the  left  support  under  joint  7.  Sometimes 
the  loads  are  varied  so  there  is  a  negative  reaction  on  the 
right,  which  means  the  end  of  the  truss  must  be  tied  down. 
Sometimes  there  is  a  small  positive  reaction  at  the  right,  which 
is  ignored  if  the  forces  are  small  in  the  members.  This  roof  is 
for  a  grand  stand.  In  a  cantilever  truss  the  loads  at  the  ends 
are  used,  whereas  they  are  ignored  in  ordinary  trusses  because 
there  they  are  carried  directly  by  the  walls. 


236 


PRACTICAL   STRUCTURAL   DESIGN 


The  reactions  are  computed  as  follows,  for  the  truss  shown, 


(300  X  36)  +  600(30  +  24  +  18  +  12  +  6) 
18 


=  3600  Ibs., 


Fig.  158  — Braced  Cantilever  with  Concentrated  Load 
at  End 


which  is  equal  to  the  total  load  on  the  truss,  therefore  the  truss 

is  exactly  bal- 
anced on  the 
support  at  joint 
5.  The  author 
has  examination 
I  V  papers  in  which 
this  roof  truss 
appears  with 
supports  at  the 
two  ends  and 
also  at  the  right 
end  and  (in 
turn)  each  ver- 
tical,  which 

gives  four  different  combinations.    Some  of  the  problems  require 

the  wind  load  to  be  computed  while  others  assume  all  loads  as 

vertical.    With  four  different  ways 

to  support  the  truss,  with   and 

without  wind  and  these  various 

conditions  varied  by  varying  the 

amount  of  vertical  load  at  each 

joint  it  is  readily  seen  why  ex- 
aminers like  this  truss.     Students 

preparing  for    architects'   license 

examinations  should  work  it  with 

a  number  of  changes  in  all  the 

conditions. 

Fig.  158    represents    a   braced 

cantilever  carrying  a  single  con- 
centrated load  at  the  end.    The 

dead  load  is  neglected  and   the 

frame  therefore  is  weightless. 
Fig.    159   represents   a   braced 

cantilever   with   a  load   at  each 

joint.     This  illustrates  the  general  method  for    dealing    with 

cantilever  trusses.     On  the  truss  diagram  the  loads  are  shown 


Fig.  159  —  Braced  Cantilever 
Loaded  on  Joints 


GRAPHIC   STATICS  237 

at  the  joints.  Multiply  each  load  by  the  horizontal  distance 
from  the  support.  Add  the  products.  Divide  by  the  sum  of 
the  loads,  to  obtain  the  position  of  the  center  of  gravity  of  all 
the  loads.  Continue  the  horizontal  reaction  line  (Ri)  to  an 
intersection  with  the,  vertical  dotted  line  through  the  center  of 
gravity.  Draw  the  diagonal  line  (#2)  to  show  the  direction  of 
the  reaction  at  the  bottom  chord  support. 

To  draw  the  stress  diagram  first  lay  off  on  a  vertical  load  line 
the  sum  of  the  loads.  From  the  upper  end  draw  a  horizontal 
line  and  from  the  lower  end  draw  a  diagonal  line  parallel  to  the 
inclined  reaction.  The  point  of  intersection,  j,  on  the  force  dia- 
gram fixes  the  amount  of  each  reaction.  From  j  drop  a  vertical 
load  line  on  which  set  off  each  joint  load  and  close  the  diagram 
at  the  bottom.  The  rest  of  the  diagram  is  evident.  The  vertical 
fe  is  not  stressed  but  is  merely  used  to  carry  the  weight  of  the 
lower  chord  in  the  end  panel. 

Accuracy  in  Drawing 

In  graphic  statics  everything  depends  on  the  care  with  which 
the  work  is  done.  The  pencils  used  should  be  very  sharp  and 
the  lines  as  thin  as  possible.  The  lines  in  the  force  diagram  must 
positively  be  parallel  with  the  lines  on  the  truss  diagram.  The 
work  checks  when  the  reciprocal  diagrams  close  and  if  they  do 
not  close  the  work  must  be  carefully  searched  for  errors.  A  use- 
ful check  is  to  determine  some  stresses  analytically  by  taking 
moments.  The  scale  should  be  one  that  will  not  require  too 
large  a  sheet  of  paper  and  will  allow  a  reading  of  one  hundred 
pounds.  For  roof  trusses  of  usual  spans  the  scale  can  be  twenty 
thousand  pounds  per  inch. 

The  foregoing  presentation  of  the  subject  of  graphic  statics 
covers  the  subject  only  so  far  as  roof  trusses  are  concerned.  It 
may  be  applied  to  any  braced  structure  and  other  applications 
will  be  shown  in  the  following  chapter.  The  principles  are  simple 
and  any  student  who  works  faithfully  through  the  examples 
given  should  have  no  hesitancy  in  attempting  to  analyze  graphi- 
cally the  forces  in  any  braced  frame. 

Continuous  Beams 

The  continuous  beam  is  not  used  much  in  steel  buildings  but 
is  used  in  all  reinforced  concrete  buildings.  All  methods  for 
dealing  with  the  continuous  beam  are  based  on  the  assumption 


238 


PRACTICAL  STRUCTURAL  DESIGN 


that  the  supports  are  at  the  same  level  and  remain  there.  If 
one  support  settles  the  stresses  are  increased  enormously.  It  is 
customary  to  use  the  "Three-moment  Theorem"  in  dealing 
with  continuous  beams,  but  it  is  rather  involved  and  a  great  many 
men  do  not  take  the  trouble  to  investigate  carefully  the  mo- 
ments on  continuous  beams.  The  following  graphical  method 
was  proposed  many  years  ago  by  T.  Claxton  Fidler  in  "A  Prac- 
tical Treatise  on  Bridge  Construction."  Another  graphical 
method  is  demonstrated  in  DuBois  " Graphical  Statics"  and 
Church's  "Mechanics  of  Engineering." 

The  Fidler  method  is  known  as  the  "  Method  of  Characteristic 
Points."    Refer  to  Fig.  44  on  page  49,  where  the  effect  of  restrain- 


(d)j-«= 

Fig.  160  —  A  Graphical  Treatment  of  Continuous  Beams.     Fidler  method 

ing  the  ends  of  a  uniformly  loaded  beam  is  discussed.  To  extend 
this  to  a  continuous  girder  uniformly  loaded  compute  the  bend- 
ing moment  on  each  span  separately  as  though  it  were  simply 
supported  at  the  ends.  The  bending  moment  =  WL  -5-  8  and  a 
parabola  must  be  drawn  on  each  span  with  the  middle  ordinate 
equal  to  the  moment,  as  in  Fig.  160. 

Divide  each  span  into  three  equal  parts  and  at  the  third  points 
erect  vertical  lines.  Make  each  vertical  line  equal  to  two-thirds 
the  height  of  the  parabola  within  which  it  is  situated.  Draw  a 
circle  around  the  end  of  the  line,  the  characteristic  point  be- 
ing in  the  center  of  the  circle.  These  characteristic  points  in 
Fig.  160  are  numbered  from  1  to  8  inclusive. 

The  end  spans,  of  the  series  here  shown,  rest  freely  on  the  outer 
end  supports,  hence  characteristic  points  1  and  8  are  disregarded. 


GRAPHIC   STATICS  239 

If  the  end  spans  were  restrained  at  the  outer  ends,  the  broken  base 
line  would  pass  through  points  1  and  8.  Starting  from  A  a  line 
is  drawn  upward  to  the  vertical  line  through  support  B.  This 
broken  line  should  pass  above  point  2  and  below  point  3,  or  it 
should  pass  below  point  2  and  above  point  3,  or  it  should  pass 
through  these  points.  In  the  present  case  it  passes  through 
them. 

Passing  through  point  3  the  line  passes  below  point  4  and 
strikes  the  vertical  line  through  support  C.  It  then  passes  above 
point  5  and  point  6  and  below  point  7  to  close  on  support  E. 

The  broken  line  must  be  fixed  by  trial  in  all  cases.  It  must 
pass  below  (or  above)  one  point  and  above  (or  below)  the  adjacent 
point  in  the  adjoining  span  to  the  vertical  line  through  the  sup- 
ports. If  the  spans  are  equal  the  broken  base  line  passes  as  far 
below  one  point  as  it  passes  above  the  adjacent  point  on  the 
adjoining  span.  When  the  spans  are  unequal  the  line  passes 
above  or  below  inversely  as  the  length  of  the  span.  That  is,  on 
the  shorter  'span  the  vertical  space  is  greater  than  it  is  on  the 
longer  span;  in  proportion  to  the  lengths  of  the  spans. 

In  fixing  the  position  of  the  broken  line  the  author  puts  over 
the  drawing  a  sheet  of  tracing  paper  on  which  to  mark  the  several 
trial  lines.  When  the  final  line  is  selected  the  points  on  the  ver- 
tical lines  through  supports  are  pricked  in  with  a  needle,  the 
tracing  paper  is  taken  off,  and  the  line  drawn.  The  line  can 
have  only  one  position. 

In  the  figure  the  moment  diagram  is  shown  at  (a).  The  broken 
line  is  a  base  line  from  which  to  scale  the  bending  moments.  All 
the  shaded  portion  within  the  parabola  on  each  span  represents 
positive  moment.  All  the  shaded  portion  outside  the  parabola 
between  it  and  the  vertical  line  through  supports  represents 
negative  moment.  At  (6)  is  shown  graphically  the  system  of 
cantilever  and  simply  supported  beams  into  which  a  continuous 
beam  over  several  supports  is  divided.  The  ends  of  the  can- 
tilevers are  at  the  point  of  contraflexure,  the  curved  line  at  (d) 
showing  the  deflection  of  the  beam  to  an  exaggerated  scale. 

At  (c)  is  the  curve  for  shears  and  also  reactions.  The  reaction 
is  always  equal  to  the  shear.  The  shear  is  zero  at  the  point  of 
maximum  bending  moment,  or,  rather,  it  passes  through  zero  at 
this  point,  the  sign  for  shear  being  positive  (+)  above  the  base 
line  and  negative  (— )  below  the  line.  The  reaction  on  either 


240  PRACTICAL   STRUCTURAL   DESIGN 

side  of  the  support  is  equal  to  the  shear  on  the  same  side.  The 
total  reaction  on  any  support  is  the  sum  of  the  positive  and  nega- 
tive shear. 

To  compute  the  shear  and  reactions  proceed  as  follows:  Shear 
at  A  (+)  =  half  the  uniform  load  on  span  1  to  2.  The  reaction 
is  equal  to  the  shear. 

Shear  on  left  of  B  (— )  =  total  load  on  the  span,  less  left  reaction. 

Shear  on  right  of  B  (+)  =  load  on  cantilever  3  to  4,  plus  half 
the  load  on  the  suspended  span  4  to  5. 

Reaction  on  support  B  =  sum  of  the  +  and  —  shear,  as  found 
above. 

Shear  on  left  of  C  (— )  =  half  the  load  on  span  4  to  5,  plus  the 
load  on  the  cantilever  5  to  6. 

Shear  on  the  right  of  C  (+)  =  load  on  cantilever  6  to  7,  plus 
hah'  the  load  on  the  suspended  span  7  to  8. 

Reaction  on  support  C  =  sum  of  the  +  and  -  shear. 

Shear  on  the  left  of  D  (-)  =  hah"  the  load  on  suspended  span 
7  to  8,  plus  the  load  on  the  cantilever  8  to  9. 

Shear  on  the  right  of  D  (+)  =  load  on  cantilever  9  to  10,  plus 
half  the  load  on  the  suspended  span  10  to  11. 

Reaction  on  support  D  =  sum  of  the  +  and  —  shear. 

Shear  and  reaction  at  E  =  half  the  load  on  the  span  10  to  11. 

Check  the  results  by  using  the  formulas  on  page  52. 

The  method  for  obtaining  moments,  shears,  and  reactions  by 
the  use  of  "characteristic  points"  may  be  used  for  any  number 
of  spans,  equal  or  unequal,  all  spans  loaded  or  some  carrying  a 
live  and  dead  load  and  others  carrying  only  a  dead  load. 


CHAPTER  VII 
Columns  and  Structures 

A     PIER  is  made  of  brick,  stone,  or  concrete.     That  is,  it  is  a 
l\  masonry  post  and  because  it  is  not  safe  to  permit  any  bending 
stress  it  must  be  limited  in  height.     A  concrete  pier  rein- 
forced with  steel  may  develop  into  a  slender  column. 

The  Chicago  Building  Ordinance  provides  that  no  masonry 
pier  can  have  a  height  exceeding  12  times  the  least  thickness. 
When  the  height  is  less  than  6  times  the  least  thickness  the  allow- 
able unit  compressive  stress  is  that  fixed  in  the  ordinance.  When 
the  height  exceeds  6  times  the  least  thickness  the  fiber  stress  must 
be  reduced  by  the  following  formula: 


in  which  /  =  reduced  unit  compressive  stress; 

c  =  unit  compressive  stress  mentioned  in  the  ordinance; 
H  =  height  in  feet; 
D  =  least  width,  or  thickness,  in  feet. 

The  distinction  between  posts  and  columns  is  seldom  definitely 
drawn.  It  may  be  said  that  a  post  is  solid  and  short.  -A  column 
is  long  and  may  be  hollow  or  of  some  shape  other  than  round  or 
rectangular. 

Specifications  vary  with  the  ideas  of  the  men  who  write  them 
and  great  differences  exist  between  specifications  and  building 
ordinances  the  country  over.  The  statements  made  in  this  sec- 
tion are  therefore  not  to  be  taken  as  meeting  the  requirements 
of  the  leading  designers  but  are  presented  merely  as  examples 
of  how  such  things  are  regulated  in  some  places. 

In  Chicago  the  maximum  length  of  timber  posts  cannot  exceed 
30  diameters,  or  30  times  the  least  thickness.  Hereafter  when 
diameter  is  mentioned  in  connection  with  columns  and  posts 
it  is  understood  to  mean  also  the  least  thickness,  if  the  column 
is  rectangular  and  not  round.  Timber  posts,  or  columns,  can- 
not be  used  in  buildings  over  one  hundred  feet  in  height,  nor  in 

241 


242  PRACTICAL   STRUCTURAL   DESIGN 

buildings  of  a  greater  height  than  twice  the  width,  for  wooden 
posts  are  not  continuous  and,  therefore,  cannot  be  relied  upon 
in  case  of  heavy  winds  to  stiffen  the  building. 

The  allowable  unit  compressive  stress  for  wood  is  for  short 
blocks  only,  in  which  the  length  does  not  exceed  twice  the  di- 
ameter. The  unit  stress  on  wooden  posts  is  found  by  the  follow- 
ing formula: 

L 


in  which/  =  reduced  fiber  stress; 

c  =  unit  compressive  stress  mentioned  in  ordinance; 
L  =  length  in  feet; 

d  =  diameter,  or  least  thickness,  in  feet; 
C  =  a  constant,  which  is  80  in  Chicago  and  has  values 
ranging  from  60  to  100  in  other  cities. 

The  above  formula  is  called  in  some  places  the  "Straight-line 
formula"  and  in  other  places  the  "Winslow  formula,"  from 
Benjamin  Winslow,  who  is  credited  with  being  the  originator. 

Wooden  posts  should  be  solid.  A  number  of  experiments 
made  on  wooden  posts  built  up  of  thick  planks  spiked  side  by  side 
showed  that  the  strength  of  such  posts  is  not  the  sum  of  the 
strength  of  the  planks.  Each  plank  when  loaded  on  the  end 
tends  to  deflect  as  though  it  were  a  long  slender  column.  Some 
experiments  made  for  Mr.  Dewell  in  California  on  short  models 
gave  better  results  than  any  other  recorded  experiments,  but 
his  built-up  posts  were  better  made  than  is  apt  to  be  the  case 
with  full-size  posts.  If  a  wooden  post  is  built  up  shear  pins  must 
be  put  between  the  planks  and  bolts  must  go  through  the  other 
way.  Sometimes  plates  on  the  edges  securely  screwed  to  each 
of  the  planks  will  make  them  act  together. 

Only  very  short  posts  fail  by  crushing  under  load.  The  usual 
failure  in  posts  and  columns  is  caused  by  bending,  which  crushes 
the  fibers  on  the  concave  side  and  frequently  causes  tension  in 
the  convex  side.  For  long  columns  it  is  necessary  to  use  a  for- 
mula for  reducing  the  allowable  compressive  stress  for  short  blocks. 
The  stress  is  progressively  reduced  as  the  length  of  the  column 
is  increased  until  finally  a  point  is  reached  beyond  which  the 
"slenderness  ratio"  is  so  great  that  the  column  may  be  unsafe. 

The  "  slenderness  ratio"  is  -.  for  masonry  piers  and  wooden  posts 


COLUMNS  AND   STRUCTURES  243 

and  it  is  -  for  metal  columns.    The  factor  "r"  is  the  "radius 

of  gyration," 

Before  describing  this  important  factor  the  general  question 
of  column  formulas  may  well  be  touched  on.  The  Euler  formula 
is  intended  for  such  long  slender  columns  that  it  is  not  in  prac- 
tical use,  being  of  value  to  investigators  and  mathematicians  in 
studying  the  effect  of  loads  applied  at  the  end  of  pieces  like 
piston  rods. 

At  least  a  century  ago  Tredgold  proposed  a  general  form  for 
column  formulas  and  this  was  later  modified  by  Professor  Gordon, 
so  it  appeared  as  follows: 


in  which  /  =  reduced  unit  fiber  stress, 

c  =  allowable  compressive  unit  stress, 

k  =  a  constant, 

L  =  length, 

d  =  diameter,  or  least  thickness. 

The  constant  "k"  depended  not  only  upon  the  material  but 
on  the  shape  of  the  section. 

With  the  Gordon  formula  it  was  necessary  to  make  innumerable 
experiments  and  thus  obtain  constants.  It  would  be  necessary 
to  make  columns  of  many  sizes  and  of  every  imaginable  shape, 
built  up  in  every  conceivable  way,  and  test  them  to  destruction 
in  order  to  be  able  to  design  similar  columns. 

Professor  Rankine,  who  succeeded  Gordon  as  Professor  of 
Civil  Engineering  in  the  University  of  Glasgow,  modified  the 
Gordon  formula  by  substituting  the  radius  of  gyration  for  the 
diameter.  A  great  many  writers  refer  to  the  Gordon  formula 
when  they  mean  the  Rankine  formula,  and  others  refer  to  the 
Rankine  formula  as  the  Gordon-Rankine.  There  appears  to 
be  considerable  confusion  as  to  what  constitutes  the  difference, 
and  some  men  do  not  appear  to  realize  that  there  is  any  difference. 

The  Rankine  formula  is  essentially  a  modification  of  the  Euler 
formula  by  combining  the  underlying  principles  of  that  formula, 
which  dealt  with  a  thread,  with  the  Gordon  formula  which  took 
into  account  the  fact  that  a  column  had  thickness  as  well  as 
length.  The  Rankine  formula  is  known  in  Germany  as  the 


244  PRACTICAL  STRUCTURAL  DESIGN 

Schwarz  formula,  an  engineer  of  that  name  on  the  continent 
of  Europe  having  developed  it  independently  of  Rankine  about 
the  same  time.  A  number  of  other  men  proposed  the  same,  or 
a  similar,  form  for  the  expression  but  Rankine  and  Schwarz  ob- 
tained the  best  publicity  in  advance  of  their  colleagues. 

In  the  Rankine  formula  the  constant  "k"  of  the  Gordon  formula, 
which  was  fixed  by  the  shape  and  the  material,  becomes  the  con- 
stant "a"  which  is  fixed  by  the  material  alone.  The  constant 
is  modified  by  the  method  of  supporting  the  ends  of  the  column: 

Rankine  formula  for  flat  ends  (fixed  in  direction) : 


1  +alr 

For  rounded  ends  (direction  not  fixed)  multiply  a  by  4. 

For  hinged  ends  (position  fixed  but  direction  not  fixed)  multi- 
ply a  by  2. 

For  one  end  flat  and  the  other  round  multiply  a  by  1.78. 

In  the  Rankine  formula  the  compressive  fiber  stress  was  the 
breaking  strength  and  the  reduced  stress  was  the  reduced  break- 
ing strength,  which  was  divided  by  a  factor  of  safety  to  obtain 
the  safe  working  stress.  The  same  result  is  obtained  by  dividing 
the  breaking  strength  by  the  factor  of  safety.  For  example 

Rankine  used   c  =  70,000  and  a  =  ^QQQ     The  stress  /  was  di- 
vided by  the  factor  of  safety  5.    To-day  c  =  14,000  and  a  = 

as  before,  but  /  is  the  safe  unit  stress  without  further  operation. 

The  values  of  the  breaking  stress  and  the  empirical  constants 
to  use  in  the  Rankine  formula  were  experimentally  determined 
by  Christie  and  Hodgkinson  many  years  ago  as  follows: 

Hard  steel,  c  =  70,000  Ibs.  per  sq.  in.     a  = 


Mild  steel,  c  =  48,000  Ibs.  per  sq.  in. 


20,000 
1 


30,000 
Wrought  iron,  c  =  36,000  Ibs.  per  sq.  in.     a 

Cast  iron,  c  =  80,000  Ibs.  per  sq.  in.     a 


Timber,  c  =    7,200  Ibs.  per  sq.  in.     a 


3000 


COLUMNS  AND   STRUCTURES  245 

The  recommended  factors  of  safety  were  as  follows:  10  for 
timber;  5  for  metal  under  moving  load;  4  for  metal  under  quies- 
cent load.  The  constants  were  for  ratios  of  -  between  20  and  200, 

and  are,  therefore,  not  reliable  for  longer  columns. 

The  Chicago  building  ordinance  limits  the  extreme  length  of 
cast  iron  columns  to  70  times  the  least  radius  of  gyration.  The 
length  of  rolled  steel  compression  members  cannot  exceed  120 
times  the  least  radius  of  gyration,  but  the  limiting  ratio  of  struts 
for  wind  bracing  may  be  150  times  the  least  radius  of  gyration. 
See  some  of  the  specifications  recommended  for  study  and  com- 
pare them  with  the  provisions  above  quoted. 

Radius  of  Gyration 

The  radius  of  gyration  was  once  humorously  referred  to  as  a 
happy  thought  in  terminology  as  it  is  not  a  radius  and  has  noth- 
ing to  do  with  gyration.  It  is  a  term  used  by  mathematicians 
and  students  of  mechanics  of  materials  to  describe  a  factor  used 
in  the  design  of  compression  members  in  structures.  It  is  ac- 
tually the  square  root  of  the  moment  of  inertia  of  a  section  divided 
by  the  area,  or, 


in  which  r  =  radius  of  gyration, 
I  =  moment  of  inertia, 
A  =  area  of  section. 

The  moment  of  inertia  and  the  area  being  in  inches,  the  radius 
of  gyration  is  in  inches. 

The  would-be  humorist  was  wrong  in  his  statement,  for  the 
radius  of  gyration  may  be  shown  to  be  a  radius  and  it  has  actually 
to  do  with  gyration. 

Each  cross  section  has  two  radii  of  gyration,  one  perpendicular 
to  the  axis  y  and  the  other  perpendicular  to  the  axis  x.  In  using 
a  formula  the  least  radius  is  chosen,  except  when  it  may  be  safe 
to  use  the  greater.  Assume  that  the  mass  rotates  (bends)  about 
the  given  axis.  If  the  column  bends,  some  resistance  will  be 
offered  by  the  section,  which  is  assumed  to  be  a  mass  moved  by 
the  rotation  (bending)  of  the  column  about  the  axis  chosen. 

Assuming  the  section  to  be  a  rotating  body,  there  is  some  kinetic 
energy  developed,  and  in  order  to  find  the  amount  it  is  first  neces- 


246  PRACTICAL  STRUCTURAL  DESIGN 

sary  to  determine  the  point  through  which  the  kinetic  energy 
acts,  or  determine  what  is  essentially  a  center  of  gravity.  The 
moment  of  inertia  is  the  sum  of  all  the  small  units  of  a  section 
multiplied  by  the  square  of  the  distance  of  each  unit  from  the 
axis.  Divide  this  by  the  area  and  extract  the  square  root  and 
the  radius  of  gyration  is  found  to  be  the  root  mean  square  of  the 
distances  of  all  the  separate  units  of  the  mass  from  the  axis.  It 
is  actually  a  radius  from  the  axis  to  the  center  of  an  imaginary 
ring  in  which  is  assumed  to  be  concentrated  the  mass  of  the  sec- 
tion. If  this  is  not  plain  the  author  offers  his  apologies,  for  he 
cannot  make  it  any  plainer  without  wandering  off  into  a  mathe- 
matical demonstration  which  would  defeat  the  objects  aimed 
at  in  writing  this  book.  It  is  of  little  consequence,  however,  as 
it  is  enough  to  accept  the  judgment  of  eminent  men  who  have 
worked  the  matter  out  satisfactorily. 

Straight-line  Formula 

The  Euler  formula  applies  only  to  the  thread  in  the  vertical 
axis  of  a  very  long  and  very  slender  column  and  the  Rankine 
formula  is  also  to  a  large  extent  a  theoretically  correct  formula, 
the  value  of  which  is  seriously  affected  by  faults  in  workmanship 
and  design,  as  well  as  by  defects  in  materials.  It  is  a  laborious 
formula  to  use  and  engineers  like  simple  formulas.  In  making 
tests  of  full-size  columns  it  was  found  by  plotting  the  results 
that  the  Rankine  formula  gives  rather  low  stresses  for  columns 
having  a  slenderness  ratio  under  80,  and  somewhat  high  stresses 
for  a  slenderness  ratio  over  150.  A  straight  line  drawn  through 
the  points,  on  the  sheet  on  which  the  results  of  the  tests  were 
plotted,  in  such  a  way  that  it  passed  through  the  center  of  mass 
of  the  points,  resulted  in  the  formula: 

/  =  16,000  -  70J;, 

which  is  known  as  the  "Straight-line"  formula  for  steel  columns. 
It  is  used  in  Chicago,  where  the  formulas  for  wrought  iron  and 
cast  iron  are  as  follows: 

wrought  iron,  /  =  12,000  -  60- 
cast  iron,  /  =  10,000  -  60^ 


COLUMNS   AND   STRUCTURES  247 

but  in  no  case  is  the  maximum  stress  permitted  to  exceed  that 
fixed  in  the  ordinance,  14,000  for  steel;  10,000  for  wrought  iron; 
and  10,000  for  cast  iron.  For  steel  columns  filled  with,  and  en- 
cased in,  concrete  extending  at  least  three  inches  beyond  the 
outer  edge  of  the  steel,  where  the  steel  is  calculated  to  carry  the 
entire  live  and  dead  load,  the  allowable  stress  per  square  inch  on 
the  steel  is  determined  by  the  following  formula, 

/  =  18,000  -  70-r, 

but  cannot  exceed  16,000  Ibs. 

The  student  is  referred  to  the  following  pages  in  the  standard 
steel  handbooks:  Carnegie  (1913),  251  to  282  incl,  327  to  329 
inch  Cambria  (1914),  192  to  276  incl.,  394-5.  Jones  &  Laugh- 
lin  (1916),  176  to  217  incl.,  281  to  283  incl.  Lackawanna  (1915), 
205  to  288  incl.,  Bethlehem  (1911),  8,  43  to  55  incl.,  70  to  87  incl., 
97  and  98.  In  addition  to  the  information  contained  on  the 
pages  mentioned  there  are  tables  of  the  radius  of  gyration  of 
pieces  having  different  shapes  and  of  pieces  in  combination, 
such  as  angles  back  to  back,  etc. 

In  Fig.  162  the  curves  show  the  allowable  fiber  stresses  per- 
mitted in  the  larger  American  cities  and  given  in  various  steel 
handbooks,  etc.  Some  of  the  curves  are  for  modifications  of  the 
original  Rankine  formula  with  the  theoretically  correct  curve 
according  to  that  formula.  Others  give  values  according  to 
various  straight-line  formulas.  Speaking  generally  the  first  figure 
in  the  straight-line  formula  gives  a  fair  idea  of  the  factor  of  safety 
intended  by  the  man  responsible  for  the  expression.  Assuming 
a  maximum  strength  of  64,000  Ibs.  per  sq.  in.  for  structural  grade 
steel,  the  factor  of  safety  when  the  formula  starts  with  16,000 
Ibs.  is  4;  for  19,200  Ibs.  it  is  3.333,  etc.  This  is  modified  again 
by  the  slope  of  the  curve  (even  straight  lines  being  called  curves 
in  graphical  work).  Notice  that  there  is  a  top  limit  when  the 
curve  goes  horizontally,  this  being  14,000  Ibs.  for  Chicago. 

To  use  the  chart  determine  the  ratio  of  slenderness  within 
which  the  column  length  is  fixed.  Assume  a  section  by  trial 
and  determine  the  radius  of  gyration  in  inches.  Divide  the 
length  in  inches  by  the  radius  of  gyration  to  obtain  the  slender- 
ness  ratio  which  must  be  within  the  limit  decided  upon.  If  it  is 
past  the  limit  the  work  must  be  done  over  with  another  assumed 
section.  If  the  slenderness  ratio  is  within  the  limit  find  it  at  the 


248 


PRACTICAL   STRUCTURAL   DESIGN 


bottom  of  the  chart  and  go  vertically  upward  to  the  curve  repre- 
senting the  formula  used.  From  this  intersection  proceed  hori- 
zontally to  the  left,  where  the  proper  fiber  stress  will  be  found. 
Divide  the  load  by  this  fiber  stress  and  obtain  the  required  cross- 
sectional  area  of  the  column.  If  it  agrees  with,  or  is  less  than, 
the  area  of  the  assumed  column  section  the  assumed  section  may 
be  used  and  the  designing  of  the  details  proceeded  with.  Another 


ch 

55 


NC| 


^ 


H?j 


S  ^  Schneider 

*KS       l~  Dr., 


Chicago  1913    Hax.lWOO 

HurWHrOK    4-— I 

Omaha  1915    \       I 
Detroit  1916  Han  12000 
Seattle  1316    »  16000 
liOOO 
»  13000. 


\  XJX©  r60M?O.I9000-IOO?\  \\      \ 


^  1      I.I 

(Indianapolis  1316 
Minneapolis     /9/f 
Cincinnati  Max.  13000 
Buffalo        "  12000 


Cam 


(D 


Cambria 


'  26000r2 


0      10     10    3P    40     50    60 


70    80    90    100 
Value  of  -V 


110   120    130    140   \SO  160    170  160   190  200 
m  Inches  > 


Fig.  162  —  Steel  Column  Formulas  Used  in  the  United  States 

way  is  to  multiply  the  cross-sectional  area  of  the  assumed  column 
section  by  the  fiber  stress  and  if  this  gives  a  carrying  capacity 
equal  to,  or  greater  than,  the  load  the  section  may  be  used.  If 
it  is  less,  then  another  section  must  be  assumed  and  the  work 
gone  through  again. 

The  steel  handbooks  contain  tables  of  columns  intended  to 
save  the  designers  much  of  the  above  work.  In  using  these  tables 
the  designer  will  notice  that  values  of  the  carrying  capacities,  for 
the  columns  are  given  with  the  lesser  and  with  the  greater  radius 
of  gyration.  Be  careful  in  using  the  tables  to  see  which  value 
is  used.  Either  value  may  be  used  in  computing  the  effect  of 
eccentric  loads,  depending  upon  which  side  the  load  comes.  The 
smaller  radius  of  gyration  is  used  in  determining  the  unit  stress 


COLUMNS   AND   STRUCTURES  249 

for  concentric  loading.  However,  there  are  cases  when  the  use 
of  the  larger  radius  of  gyration  may  be  permitted.  If  a  column 
is  built  into  a  wall  of  first-class  masonry  so  that  it  cannot  bend 
in  the  weaker  direction  the  larger  radius  of  gyration  may  be 
used.  A  casing  of  a  few  inches  of  concrete  is  not  enough  to  satisfy 
the  requirement  that  the  column  be  stayed  in  the  weaker  direc- 
tion. The  supported  length  of  a  column  is  the  length  used  in 
the  formulas.  If  a  column  is  supported  in  the  weaker  direction 
by  adequate  bracing  the  supported  length  is  the  distance  between 
the  attached  ends  of  the  stays,  and  the  column  may  be  designed 
with  the  smaller  radius  of  gyration  combined  with  the  shorter 
lengths,  or  it  may  be  designed  with  the  larger  radius  of  gyration 
combined  with  the  greater  length.  When  possible  the  weaker 
dimension  of  the  column  should  be  turned  in  the  direction  of 
the  closer  supports.  Even  when  the  least  radius  of  gyration  is 
chosen  the  column  should  be  so  placed  in  the  structure  that  the 
heavier  loads  come  on  the  longer  axis. 

The  effect  of  eccentric  loading  is  taken  care  of  by  increasing 
the  size  of  the  column.  The  tendency  of  the  column  to  bend  is 
determined  by  the  slenderness  of  the  section  and  it  can  bend 
sideways  to  the  load,  this  being  the  reason  for  using  the  least 
radius  of  gyration  regardless  of  the  direction  from  which  the  load 
may  come  to  the  column.  In  the  column  tables  in  the  steel  hand- 
books the  total  load  is  generally  given,  together  with  a  state- 
ment as  to  which  radius  of  gyration  is  used  in  computing  the 
strength  of  the  column.  The  tables  are  computed  by  one  of  the 
several  formulas  plotted  in  Fig.  162. 

In  assuming  column  sections  the  formulas  given  do  not  take 
into  account  the  various  methods  for  attaching  the  principal 
parts  together.  From  the  result  of  experiments  it  is  believed  safe 
to  use  the  allowable  fiber  stress  by  formula  for  columns  with 
solid  web  plates,  as  for  example  plate  and  angle  columns.  For 
laced  columns  use  about  seventy-five  per  cent  and  for  columns 
fastened  by  batten  plates  use  about  fifty  per  cent  of  the  fiber 
stress  given  by  formula. 

Fig.  163  appeared  in  Engineering  News,  in  1913  in  an  article 
by  O.  von  Voigt lander  on  Approximate  Radii  of  Gyration.  The 
use  of  the  table  saves  a  great  deal  of  labor  on  the  part  of  the 
designer  when  he  can  know  in  advance  the  outside  dimensions 
of  his  columns  or  struts.  It  is  not  necessary  to  know  this  accu- 


250 


PRACTICAL   STRUCTURAL   DESIGN 


k~b-- 


Y 
F"     "1 


13 


Y 

!  JTT  r= 


x   0.31  h 

w 

25 


h=mean  h 
2 


J  i 


r  Same  as 
*J   No.7,6,9 


14 


AT 

^^ 


rx=  0  15  h 


U- 


^  rx=  0.40  h 
H  ry=  0.11  b 


29 


18 


rx=0.38h 

ry=  °-2Zb 
30 


h~rx=  0.31  h 

T-ry=  0.215  b 


bmus+beabf. 


H —  b  — H 

ir 


0.21  b 


MTrx= 
hr9 
^ 


0.36  h 
0.53b 


rx=  0.39  h 
r 


0.435  h 
rq>0.25  b 
-ib  33 


"^hf 
—  i 


-r«=  0.41  h 


35 


l/      rx=  O.ZIh 
~ma*—i\    0.2!  b 


__ry= 


12 


-0.26b 
14 


b=  Width  of  Section 
ParaUeltoAxis  X-X 
h=  Height  of  Section 
Parallel  to  Axis  Y-Y 


Fig.  163  —  Approximate  Radii  of  Gyration  "r" 


COLUMNS  AND   STRUCTURES  251 

rately,  but  usually  it  can  be  determined  in  advance  just  what 
maximum  size  is  permissible.  In  the  figure  the  meaning  of  the 
letters  used  is  plain,  but  attention  must  be  called  to  19  in  which 
"b"  is  the  distance  back  to  back  of  the  channels  and  it  must  be 
not  less  than  63  per  cent  of  the  nominal  size  of  the  channel.  The 
procedure  is  to  assume  the  form  of  section  and  the  extreme 
dimensions.  Then  apply  the  rules  given  in  Fig.  163  and  thus 
get  an  approximate  value  for  the  radius  of  gyration.  Proceed  as 
before  and  when  the  allowable  fiber  stress  is  found  proceed  to 
get  the  area  and  then  select  the  plates  and  shapes  to  make  the 
selected  section.  When  it  has  been  designed  find  the  exact 
radius  of  gyration  and  test  for  the  fiber  stress. 

Wrought  iron  columns  are  seldom  used,  for  the  material  is 
hard  to  obtain  and  steel  is  stronger  pound  for  pound.  Wrought 
iron  and  steel  columns  are  usually  two  or  three  stories  long. 
Column  splices  should  be  so  arranged  that  not  more  than  one- 
half  the  total  number  of  columns  splice  at  any  one  floor  level. 
All  connections  between  columns,  girders,  and  beams  should  be 
ri vetted.  Theoretically  it  is  best  to  vary  the  sizes  of  columns 
from  story  to  story,  but  it  is  less  expensive  with  steel  and  wrought 
iron  columns  to  have  them  not  less  than  two  stories  long,  of  the 
same  size,  for  the  extra  amount  of  material  often  costs  less  than 
the  labor  required  to  change  sizes  at  each  floor.  Cast  iron  columns 
and  wooden  columns  are  never  more  than  one  story  in  length 
and  the  practical  impossibility  of  making  rigid  connections  at 
floor  levels  limits  the  use  of  cast  iron  and  wood  for  columns  to 
low  buildings,  for  they  offer  poor  resistance  to  wind. 

No  column  is  free  to  turn  as  though  the  end  were  round  or 
as  if  it  bore  against  a  pin.  Such  conditions  do  not  arise  in  build- 
ing construction,  although  they  may  be  nearly  attained  in  bridges. 
The  columns  in  massive  buildings  are  sometimes  considered  as 
fixed  at  the  ends,  but  mass  implies  positive  rigidity.  In  sheds  and 
low  mill  and  shop  structures  columns  are  not  considered  as  fixed 
at  the  ends  unless  specially  massive  foundations  are  used  for 
the  purpose  of  assuring  such  a  condition,  something  seldom 
done.  The  majority  of  engineers  advocate  the  assumption 
of  two  rounded  ends  for  all  cases  short  of  positive  fixety,  as 
there  are  so  many  secondary  stresses,  experiments  showing 
that  columns  tested  to  destruction  fail  in  detail  rather  than  as 
a  whole. 


252 


PRACTICAL   STRUCTURAL   DESIGN 


Fig.  164  illustrates  the  four  conditions  affecting  the  end  load- 
ing of  columns.  At  (a)  is  shown  the  column  with  pin-joints  at 
top  and  bottom.  This  is  the  standard  case  assumed  for  all  column 
formulas,  a  modification  being  a  flat-ended  column,  the  bolts  at 
the  ends  of  which  are  intended  to  merely  hold  it  in  position  and 
are  not  strong  enough  to  resist  much  tension.  The  I  in  the  for- 
mulas is  the  total  unsupported  length  of  the  column. 

At  (6)  is  a  column  fixed  at  the  ends  to  maintain  both  position 
and  direction.  The  I  to  use  is  one-half  the  unsupported  length. 

At  (c)  one  end  is  fixed  (in 
position  and  direction)  and 
the  other  end  is  a  pin,  or 
hinged,  end,  fixed,  in  position 
but  not  in  direction.  The  I 
to  use  is  two-thirds  the  un- 
supported length. 

At  (d)  the  lower  end  of  the 
column  is  fixed  in  position 
and  direction  but  the  upper 

end  is  free  to  move  laterally, 

Fig.  164  —  Methods  of  Fixing  Columns     ,  .  f 

differing  from  (a)  and  (c)  in 

which  the  upper  end  of  the  column  is  vertically  over  the  lower 
end.  The  column  shown  at  (d)  bends  in  a  simple  curve  which  is 
one-half  that  of  a  column  double  the  length,  as  shown  by  the 
lower  dotted  end.  Obviously  the  I  to  use  is  twice  the  actual 
length  of  the  column. 

The  temptation  to  consider  a  greater  degree  of  fixity  than  is 
actually  obtained  is  great,  and  all  designers  must  be  warned 
against  yielding  to  it.  Probably  the  only  fixed  columns  in  a  build- 
ing are  those  on  the  ground  floor  of  a  high  building  with  massive 
foundations.  Above  the  ground  floor  there  is  bound  to  be  some 
vibration  and  swaying,  especially  in  a  wind. 

To  Proportion  Struts  or  Compression  Members 

Every  strut  is  designed  as  a  column  with  rounded  or  hinged 
ends.  Nothing  is  deducted  for  rivet  holes,  as  the  rivets  are  assumed 
to  fill  them.  First  select  the  form  of  strut  from  the  many  illus- 
trated in  Fig.  163,  decide  on  the  radius  of  gyration,  and  be  careful 
with  angles  to  use  two,  back  to  back,  even  when  the  computa- 
tions show  one  to  be  amply  strong.  Steel  compression  members 


COLUMNS   AND   STRUCTURES 


253 


71 


1  w 


Fig.  165  —  Eccentric  Loads  on  Columns 


in  trusses  are  apt  to  contain  considerable  excess  material  because 
they  are  usually  composed  of  angles,  but  the  excess  material  is 
often  of  considerable  advantage  when  wind  is  considered.  The 
consideration  of  holes  in  tension  members  leads  to  an  excess  of 
material,  and  the  effect  of  the  radius  of  gyration  is  similar  in  com- 
pression members. 

Eccentric  Loads  on  Columns 

The  column  formulas  heretofore  considered  are  based  on  a 
load  acting  vertically  and  applied  at  the  upper  end  of  the  vertical 
axis  of  the  column.  This  is  termed  concentric  loading.  The  only 
tendency  to  bend  is  that  caused 
by  the  fibers  being  too  strong 
to  crush  or  tear  until  after 
considerable  side  bending 
takes  place. 

An  eccentric  load  is  one  ap- 
plied at  some  distance  off  the 
center  of  the  column  and  act- 
ing vertically.  This  is  illustrated  in  Fig.  165.  At  (a)  two  loads 
are  shown  carried  on  opposite  sides  of  the  column.  Each  reaction 
is  assumed  to  act  vertically  at  the  middle  of  the  bracket  on  which 
the  beam  rests.  From  the  reaction  of  A  to  the  axis  of  the  column 
the  distance  is  x  and  from  the  reaction  of  B  to  the  axis  of  the 
column  the  distance  is  y.  The  center  of  gravity  of  the  two  loads 
is  found  by  the  principle  of  momentSjjmd  the  distance  of  the 
center  of  gravity  from  the  axis  of  the  column  is  e^  the  eccentri- 
city of  the  loads. 

(Ax)  +  (By) 
A  +  B 

This  eccentricity  is  always  on  the  side  of  the  heavier  load. 

At  (6)  the  eccentricity  is  the  distance  from  the  center  of  the 
bracket  support  to  the  center  line  (axis)  of  the  column.  In  both 
cases  the  load  is  the  total  eccentric  load,  which,  multiplied  by 
the  eccentricity,  causes  a  bending  moment  in  the  column.  This 
bending  moment  increases  the  compression  on  the  side  of  the 
column  on  which  the  eccentricity  exists  and  causes  tension  on  the 
far  side.  Sometimes  this  tension  may  be  great  enough  to  over- 
come the  compression  caused  by  the  direct  load,  in  which  case 
the  column  will  fail.  All  columns  are  loaded  eccentrically  and 


254  PRACTICAL   STRUCTURAL   DESIGN 

suggested  methods  for  dealing  with  this  condition  are  not  always 
correct. 

The  effect  of  an  eccentric  load  is  assumed  to  disappear  at  each 
story  height.  That  is,  the  tension,  or  additional  compression, 
caused  in  any  story  length  of  column  by  eccentric  loading  is 
assumed  to  be  a  maximum  at  the  mid-length  of  the  column  on 
that  story  and  to  be  zero  at  the  ends. 

With  a  concentric  load  the  unit  compressive  stress  over  the 
cross-sectional  area  of  the  column  is 

f     W 

f  =  T 

in  which  /  =  unit  compressive  stress  in  pounds  per  square  inch, 
W  =  concentric  load  in  pounds, 
A  =  area  of  cross  section  in  square  inches. 
Let  e  =  eccentricity  in  inches, 

n  =  distance  from  axis  of  column  to  extreme  fiber  in  the 

direction  of  the  eccentric  load, 
P  =  eccentric  load  =  A  or  A  +  B. 

The  bending  moment  due  to  the  eccentric  load  is 

M  =  Pe, 

and  the  extreme  fiber  stress  due  to  the  combination  of  direct  and 
eccentric  load  is 


in  which  /  =  moment  of  inertia  in  direction  of  bending. 

The  positive  sign  (+)  is  used  to  obtain  the  compression  and 
the  negative  sign  (-)  is  used  to  obtain  the  tension.  The  com- 
pressive stress  cannot  exceed  the  safe  allowable  stress  determined 
by  a  column  stress  reduction  formula  and  the  tensile  stress  is 
not  considered. 

The  foregoing  is  an  approximation  only,  but  is  satisfactory 
when  the  flexural  stress  due  to  eccentric  loading  is  not  large. 
The  majority  of  designers  use  only  seventy-five  per  cent  of  the 
moment  due  to  eccentricity  and  reduce  the  eccentric  load  to  an 
equivalent  concentric  load  by  the  following  expression: 

We  =  0.75  (^ 

in  which  We  =  the  equivalent  eccentric  load, 
r  =  radius  of  gyration. 


COLUMNS   AND   STRUCTURES  255 

With  the  both  formulas  it  is  necessary  to  select  a  column  sec- 
tion and  obtain  the  values  of  e,  r  and  n.  There  is  little  difference 
in  the  values  of  r  and  n  between  columns  of  nearly  the  same  size, 
so  one  computation  for  the  value  of  We  will  generally  be  sufficient. 
An  experienced  designer  can  usually  select  a  trial  size  so  nearly 
right  that  but  one  approximation  will  be  necessary. 

Having  obtained  We  it  is  necessary  to  add  to  it  the  direct  con- 
centric load  W  and  the  eccentric  load  P;  thus  WeW P  =  Wt  the 
.total  equivalent  concentric  load,  and  the  uniform  compressive 

fiber  stress  becomes  /  =  -p 

When  something  better  than  a  very  close  approximation  (good 
enough  for  ninety-five  per  cent  of  columns)  is  wanted,  the  follow- 
ing formula  by  Professor  J.  B.  Johnson  may  be  used. 

Mn 
f*  =  -pj? 

1  WE 

in  which  fb  =  unit  flexural  stress  in  pounds  per  square  inch, 
L  =  length  of  the  piece  in  inches, 
E  =  modulus  of  elasticity  of  the  material. 

The  Johnson  formula  is  used  for  beams  subjected  to  bending 
as  well  as  to  direct  compression  or  tension;  and,  also,  to  struts 
and  ties  eccentrically  loaded  in  addition  to  having  a  concentric 
load  to  carry.  The  unit  flexural  stress  must  be  added,  algebrai- 
cally, to  the  direct  stress  due  to  the  concentric  load,  and  the  sum 
of  the  two  cannot  exceed  the  safe  fiber  stress  of  the  column  as 
determined  by  a  column  stress  reduction  formula. 

For  columns,  ties  and  struts,  the  load  P  acts  parallel  to  the 
piece  and  M  =  Pe.  For  beams  the  moment  M  is  the  bending 
moment  caused  by  the  dead  load  of  the  beam  plus  whatever  ad- 
ditional transverse  load  there  may  be  on  it.  Therefore,  for  beam 
subjected  to  direct  tension  or  compression  in  addition  to  cross 

W 

bending  the  P  is  really  W,  as  used  in  the  expression  -j-»  and  is 

A. 

not  an  eccentric  load,  but  is  the  direct  concentric  tension  or 
compression. 

The  beams  carried  by  wooden  columns  and  steel  columns  rest 
on  brackets  attached  to  the  columns.  There  can,  therefore,  be 
no  uncertainty  as  to  the  amount  of  eccentricity,  e.  Concrete 
columns  are  cast  integrally  with  beams  and  slabs,  so  considerable 


256  PRACTICAL   STRUCTURAL  DESIGN 

uncertainty  often  exists  in  the  minds  of  draftsmen  as  to  the 
amount  of  eccentricity.  The  author  assumes  that  the  load  from 
the  beam  is  delivered  to  the  column  in  a  uniformly  varying  amount 
from  the  face  of  the  column  to  the  vertical  axis.  The  vertical 
load  acting  through  the  center  of  gravity  makes  the  moment 
arm  for  each  beam  equal  to  one-third  the  width,  measured  from 
the  center  of  the  column,  that  is,  two-thirds  of  half  the  width. 
Multiply  each  load  by  this  arm,  add  the  products,  and  divide  by 
the  sum  of  the  loads,  which  will  give  the  eccentricity  measured 
from  the  center  of  the  column. 

When  columns  are  connected  by  girders  then  the  deflection  in 
the  frame  will  vary  with  the  relative  rigidities  of  the  connected 
members  and  this  will  fix  the  stresses  at  the  connections. 

Wind  Bracing  for  Columns  and  Frames 

In  the  handbook  of  the  Passaic  Rolling  Mill  Co.,  a  number  of 
years  ago  the  whole  subject  of  wind  bracing  in  buildings  was  dis- 
posed of  with  the  presentation  of 


~T    "Bethlehem  Handbook"  (1908)  of 
A    which  only  the  one  edition,  now 
r  out  of  print,  was  issued  the  same 

figures  and  formulas  were  given. 
In  the  following  formulas  col- 
umns are  considered  as  fixed  at 
both  ends.     If  columns   are   not 
!     fixed  at  the  ends  substitute  2h  for 
h,  everywhere  in  the  formulas.    All 
Fig.  166— Case  I  of  Portal  Framing   members  are  constructed  to  resist 

tension  (-)  and  compression  (+). 
H  =  total  horizontal  force  at  top  of  frame. 

Stress  in  the  knee  braces  =  ±  H  f  -=  +  —  W- 

Stress  in  the  columns  =  ±  H  i a  +-}-;• 

Stress  in  the  girder  =  ±  H  (l  +  ^ 

Mb  on  the  columns  =  H  +  -r 


Mb  on  the  girder  =  H  I ^  ~  y  (a  +  2 


xxxx 


"T 

a 


y 


COLUMNS   AND   STRUCTURES  257 

Stress  in  AB  =  =*=  H  (l  +  — 

Stress  in  CD  =  ±  H  ( •=  +  -r- 
\2      4a> 

Stress  in  diagonals  =  ±  -^  (I  +  4)  /«' 

Stress  in  columns  =  =±=  #  (a  +  -W- 

ft 

Af b  on  columns  =  H  x  T- 
4 

Diagonal  bracing  is  the  cheapest  in  tall  buildings,  but  it  is  not 
possible  always  to  use  it  on  account  of  window  or  other  openings. 
It  is,  therefore,  necessary  in  most 
cases  to  use  girders  at  floor  levels, 
as  illustrated  in  Fig.  166,  or 
trusses,  as  illustrated  in  Fig.  167. 
The  figures  show  but  one  frame 
and  apply  to  single  story  build- 
ings. It  is  easy  to  say  that  a 
high  building  may  be  considered 
to  be  a  series  of  such  frames 
superimposed  and  side  by  side.  |H 

Difficulties  arise  when  it  is  neces-       *   -  z 
sary  to  apportion  the  wind  load        Fig.  167  —  Case  II  of  Portal 

,  ,        ,  Framing 

on  each  story  and  on  each  column. 

It  is  assumed  that  at  each  floor  level  the  stiff  floors  will  distribute 
the  load  to  the  columns  and  it  is  only  in  the  vertical  frames  that 
attention  must  be  paid  to  proportioning  of  members  to  resist  wind. 

Every  building  having  a  height  twice  the  width  must  be  pro- 
portioned to  resist  wind  and  the  pressure  per  square  foot  of  the 
wind  is  fixed  by  the  ordinance  followed  by  the  designer.  If, 
there  is  no  building  ordinance  to  be  followed  then  the  usual  require- 
ment is  30  Ibs.  per  sq.  ft.  of  exposed  surface. 

The  designer  has  a  choice  of  lines  through  which  the  wind 
force  may  be  carried  to  the  foundations.  Look  up  the  specifica- 
tions and  see  just  how  the  stresses  are  fixed  when  wind  is  included. 
Then  test  each  line  of  columns  to  see  if  they  can  carry  wind  as 
well  as  the  gravity  loads.  If  they  can  do  this  the  building  is 
stable  against  wind.  If  they  will  not  do  it  the  designer  must  fix 
on  some  lines  of  columns,  with  their  connecting  floor  girders, 


H 


258  PRACTICAL   STRUCTURAL   DESIGN 

which  must  be  designed  as  a  frame  to  carry  the  wind  loads.  If, 
by  slightly  deepening  the  girders  on  all  lines  of  columns  the 
matter  can  be  accomplished,  then  all  the  lines  of  columns  may  be 
called  into  service.  Usually,  however,  the  necessity  for  making 
openings  through  interior  walls  requires  that  a  minimum  of  depth 
be  used  in  girders  and  that  the  sizes  of  columns  must  be  a  mini- 
mum. When  it  is  decided  to  keep  the  area  between  columns  as 
free  as  possible  from  obstructing  beams,  ties,  and  struts,  it  will 
be  necessary  to  select  a  few  lines  of  columns  parallel  with  the 
wind,  which,  with  their  connecting  girders,  trusses,  or  ties,  will  be 
designed  as  frames  to  resist  the  force  of  the  wind.  Wall  columns 
are  usually  chosen  and  the  spandrel  beams  are  deepened,  be- 
cause, being  in  the  walls,  they  cannot  be  in  the  way  of  partitions 
or  alterations  in  the  interior  of  the  building.  When  necessary  to 
strengthen  framework  across  the  interior  of  a  building  it  is  usual 
to  do  it  on  the  wall  lines  of  light  wells. 

For  wind  alone  the  columns  of  a  building  may  be  considered 
to  have  two  fixed  ends,  except  the  columns  supporting  the  roof. 
If  the  footings  are  not  designed  to  resist  the  additional  force  of 
the  wind  the  lower  columns  are  not  considered  fixed.  It  is  hardly 
probable  that  the  foundations  will  not  be  so  designed.  Assum- 
ing the  columns  to  be  fixed  and  the  trusses  or  girders  connecting 
the  columns  to  be  strongly  attached  to  them,  there  will  be  a  point 
of  contraflexure  in  each  column  and  in  each  girder.  For  con- 
venience in  designing  this  point  of  contraflexure  is  taken  to  be 
in  the  middle.  It  is  not  an  accurate  assumption,  but  it  is  safe 
and  lessens  the  tune  required  for  computation  and  simplifies  the 
work.  There  is  moment  only  at  the  ends,  and,  so  far  as  the  wind 
force  is  concerned,  the  columns  and  girders  can  be  hinged  at  the 
points  of  contraflexure.  This  really  means  that  each  column 
consists  of  two  cantilevers  extending  upward  and  downward 
from  the  floor  beam  with  a  length  equal  to  half  the  story  height; 
and  each  girder  consists  of  two  cantilevers  extending  to  the  right 
and  left  of  the  column  and  with  a  length  equal  to  one-half  the 
span. 

For  each  story  the  total  amount  of  wind  on  the  story  is  assumed 
to  be  concentrated  at  the  middle  of  the  column  on.  This  force 
is  horizontal  shear.  Assuming  the  tall  building  to  be  a  vertical 
cantilever  beam  the  wind  loads  are  the  loads  (horizontal  shear) 
in  each  story.  Adding  these  loads  from  the  top  down  the  total 


COLUMNS   AND   STRUCTURES  259 

shear  at  each  story  height  is  found,  precisely  as  shear  is  deter- 
mined for  any  cantilever  beam. 

This  shear  is  distributed  across  the  frame  in  the  direction  of 
the  wind  by  dividing  the  total  shear  at  any  floor  by  twice  the 
number  of  panels.  (Number  of  columns  less  1  =  number  of 
panels.)  Each  end  column  carries  the  amount  thus  found  and 
each  intermediate  column  carries  double  this  amount,  because 
the  end  columns  support  only  one-half  a  panel  and  each  inter- 
mediate column  supports  a  full  panel. 

For  the  top  story  the  formulas  in  relating  to  Fig.  166  or  Fig. 
167  may  be  used.  The  direct  stress  in  the  column  is  added  to 
the  concentric  load  in  that  column,  but  is  not  carried  to  the  floor 
below.  The  bending  moment  in  the  column  is  treated  as  a  mo- 
ment due  to  eccentric  loading. 

For  each  story  below  the  top  the  moment  for  each  column  is 
equal  to  the  total  horizontal  shear  on  that  column  multiplied 
by  the  story  height.  The  wind  force  is  assumed  to  act  at  the 
mid-height  and  as  the  column  is  practically  a  cantilever  with  a 
length  equal  to  half  the  story  height  the  proper  length  to  use  for 
this  condition  of  a  column  fixed  at  the  bottom  and  free  at  the 
upper  end  is  twice  the  actual  length. 

The  bending  moment  in  a  girder  is  in  all  cases  the  mean  be- 
tween the  bending  moments  in  the  column  below  and  above  the 
girder.  It  is  independent  of  the  span.  The  moments  in  columns 
and  girders  are  at  the  ends,  the  force  in  the  middle  being  shear. 
The  bending  moments  in  girders  must  be  provided  for  by  adding 
haunches  to  them  instead  of  using  simple  brackets.  Brackets 
on  which  girders  rest  are  assumed  to  resist  vertical  shear  by  the 
rivets  which  connect  them  to  the  columns.  They  should  be  de- 
signed, when  possible,  so  none  of  the  rivets  will  be  in  tension. 
Gusset  plates  and  brackets  for  connecting  wind-bracing  girders 
to  columns  are  in  compression  below,  or  above  the  girder  and 
are  in  tension  above,  or  below,  the  girder.  It  is,  therefore,  neces- 
sary to  use  a  very  low  tensile  stress  in  the  rivets,  which  it  is  cer- 
tain they  can  withstand,  and  this  it  will  be  found  calls  for  a 
great  number  of  rivets.  A  tensile  stress  of  6000  Ibs.  per  sq.  in. 
is  used  in  such  cases. 

Fig.  168  is  a  graphical  representation  of  the  moments  in  some 
of  the  columns  and  girders  of  a  frame.  The  moment  is  a  maxi- 
mum at  the  end  and  varies  uniformly,  so  it  is  only  necessary  to 


260 


PRACTICAL   STRUCTURAL   DESIGN 


plat  each  moment  on  opposite  sides  of  the  members  and  draw  a 
straight  line  connecting  the  end  lines.  The  shear  is  constant,  as 
it  is  a  concentrated  load. 

The  author  has  here  presented  the  method  he  uses  in  designing 

frames  of  buildings  to 
resist  wind.  There  are 
several  other  methods 
in  use,  and  not  all 
engineers  will  agree 
with  the  method  here 
given.  It  is,  however, 
simple  and  agrees  well 
with  such  meager 
knowledge  as  we  now 
possess  of  the  actual 
force  of  the  wind  on 
tall  buildings.  It  is 


Moments  in  Girders. 


Shear  on  Girders. 


Fig.  168  —  Graphic  Representation  of  Moments    probably  in  more 


common  use  than  any 


and  Shears  in  Frame  of  a  Building 

other  method.    Another  method  is  to  assume  the  building  frame 
as  a  vertical  cantilever  beam  loaded  at  the  mid-point  of  each 
story  with  the  wind  as  a  concentrated  load.     The  total  moment 
is  found  for  the  beam  at  each  floor  level 
and  this  is  divided  among  the  lines  of 
columns  proportionately  to   their   dis- 
tance from  the  neutral  axis,  which  is 
assumed  to  be  midway  between  the  ex- 
terior columns.     The  bending  moments 
in  columns  and  girders  are  equal  at  the 
respective  floor  levels. 

In  Fig.  169  is  shown  the  method  of 
bracing  frames  by  means  of  diagonal 
ties,  known  as  sway  bracing.  It  is  as- 
sumed that  the  bracing  is  provided  only 
between  the  exterior  columns  and  the 
first  line  of  interior  column,  The  wind 
load  is  applied  at  each  floor  level.  The 
two  lines  of  columns  become,  respectively,  the  upper  and  lower 
chord  of  a  cantilever  truss,  the  floor  beams  or  girders  are  the 
verticals,  and  the  diagonal  ties  carry  shear.  Such  a  method 


COLUMNS   AND   STRUCTURES 


261 


causes  no  bending  in  the  columns  but  does  add  a  direct  load.  It 
adds  a  direct  load  to  the  floor  beams  or  girders,  which  must  be 
investigated.  The  forces  may  be  ascertained  analytically  or 
graphically.  It  is  usually  best  to  put  in  counters,  as  indicated 
by  the  dotted  lines. 

Loads  on  Columns  in  Buildings 

The  dead  weight  of  a  building  is  a  constant  matter.  Live 
loads  vary  from  time  to  time.  Just  what  proportion  of  live  load 
should  be  carried  to  columns  is  not  settled,  but  it  is  common  prac- 
tice to  design  the  columns  supporting  the  roof  for  the  full  dead 
and  live  roof  load.  The  columns  supporting  the  floor  next  to  the 
roof  are  designed  to  carry  the  load  transmitted  to  them  by  the 
roof  columns  together  with  the  dead  load  of  the  floor  and  part 
of  the  live  load.  In  Chicago  only  85  per  cent  of  the  live  load  on 
the  top  floor  is  carried  to  the  columns.  In  other  cities  90  per  cent 
is  used  and  some  engineers  use  95  per  cent  of  the  live  load.  It 
is  only  on  the  top  floor  columns  that  any  difference  of  opinion 
exists.  On  all  columns  below  the  top  floor  the  live  load  is  reduced 
progressively  5  per  cent  per  floor  until  the  reduction  amounts 
to  50  per  cent  of  the  live  load.  From  this  floor,  only  50  per  cent 
of  the  live  load  on  each  floor  is  carried  to  the  column,  together 
with  the  total  dead  load.  Using  Chicago  requirements  and 
assuming  a  ten-story  building  with  100  Ibs.  per  sq.  ft.  live  load 
on  each  floor;  roof  live  load  30  Ibs.,  etc.,  the  load  per  square  foot 
carried  to  the  columns  will  be  as  follows: 


Columns 
under 

Loads  per  sq.  ft.  to  columns 
for  each  floor 

Total  load 
per  sq.  ft. 
on  column 

Dead 

Live 

Total 

Roof 

60 

30 

90 

90 

10th  floor 

75 

85 

160 

250 

9th    „ 

75 

80 

155 

405 

8th 

75 

75 

150 

555 

7th 

75 

70 

145 

700 

6th 

75 

65 

140 

840 

5th 

75 

60 

135 

975 

4th 

75 

55 

130 

1105 

3rd 

75 

50 

125 

1230 

2nd 

75 

50 

125 

1355 

1st 

75 

50 

125 

1480 

262  PRACTICAL   STRUCTURAL  DESIGN 

The  loads  are  tabulated  as  in  the  above  table  and  the  area  of 
floor  served  by  a  column  is  multiplied  by  the  total  load  per  square 
foot  shown  in  the  last  column,  in  order  to  determine  the  column 
load  at  each  floor. 

The  area  of  floor  served  by  interior  columns  is  equal  to  the 
space  enclosed  between  four  columns.  Side  wall  columns  serve 
an  area  equal  to  one-half  this  and  corner  columns  serve  an  area 
one-fourth  that  of  the  space  between  four  columns.  In  other 
words  loads  go  to  the  nearest  support.  The  wall  columns  carry 
the  additional  dead  load  of  the  walls,  each  column  carrying  a 
length  of  wall  measured  midway  between  adjacent  columns,  that 
is,  a  panel  length. 

For  the  load  permitted  on  various  soils  and  for  the  amount 
of  live  load  to  be  carried  to  foundations,  with  and  without  pil- 
ing, consult  the  various  specifications  mentioned  and  the  steel 
handbooks. 

The  distribution  of  load  on  footings  is  not  treated  adequately 
in  many  textbooks.  The  dead  load  is  constant  and  the  live  load 
is  variable.  If  the  total  load,  dead  plus  reduced  live  load,  is  used 
in  proportioning  the  footings  the  footings  under  the  interior 
columns  will  be  entirely  too  large  if  the  building  stands  vacant. 
The  footings  under  the  walls,  however,  will  continue  to  settle 
and  old  buildings  with  humps  in  the  floors  over  the  girders  were 
no  doubt  so  designed  that  the  footings  under  all  columns  were 
proportioned  for  the  total  dead  and  live  load,  or  total  dead  and 
reduced  live  load. 

A  common  method  is  to  use  for  interior  columns  the  allowable 
soil  load  and  for  wall  columns  a  soil  load  about  500  Ibs.  per  sq. 
ft.  less,  and  then  proportion  the  footings  for  the  total  load  brought 
down  as  illustrated.  Some  men  make  the  soil  load  for  the  exterior 
columns  one-third  less  than  that  for  the  interior  columns  and 
design  the  footings  for  the  total  dead  plus  the  reduced  live  load. 

Mr.  Schneider  recommends  the  following  method:  Proportion 
the  footing  under  the  column  carrying  the  maximum  live  load. 
Divide  the  total  load  by  the  allowable  soil  pressure  and  obtain 
the  square  feet  required  to  carry  the  load.  Divide  the  dead  load 
by  the  area  thus  found  and  obtain  a  reduced  soil  pressure  per 
square  foot.  Using  this  reduced  soil  pressure  design  the  rest  of 
the  footings  for  the  dead  load  only.  This  is  very  conservative 
but  may  well  be  used  for  reinforced  concrete  buildings,  as  all  the 


COLUMNS   AND   STRUCTURES  263 

beams,  girders,  and  slabs  are  designed  as  continuous.    Any  settle- 
ment will  be  bad  and  generous  foundations  will  prevent  settlement. 

The  Schneider  method  calls  for  expensive  foundations,  and 
Mr.  Daniel  E.  Moran  suggests  using  one-half  the  probable  maxi- 
mum live  load,  instead  of  the  full  live  load  advised  by  Mr.  Schnei- 
der. Mr.  Moran  says:  "The  maximum  probable  load  is  the 
load  which  in  the  opinion  of  the  designer  will  actually  come  upon 
the  footings,  and  is  to  be  determined  by  a  study  of  the  conditions 
which  will  obtain  when  the  building  is  occupied.  For  instance, 
in  a  schoolhouse  the  number  of  children  in  each  class  room  and 
the  weight  of  desks,  chairs,  etc.,  may  be  determined  with  con- 
siderable accuracy  and  these  loads  will  make  the  maximum 
probable  live  load.  As  a  further  illustration,  in  many  school- 
houses  there  is  an  assembly  room  which  is  only  used  when  the 
class  rooms  are  vacant,  and  consequently  if  class  room  loads  are 
used  assembly  room  loads  should  be  omitted  or  vice  versa,  the 
greater  one  of  these  loadings  to  be  used  for  the  probable  load." 
See  on  this  point  Engineering  News,  March  6,  1913,  and  April 
3,  1913. 

The  author  was  taught,  thirty  years  or  more  ago,  to  propor- 
tion the  loads  as  follows:  the  dead  plus  the  reduced  live  loads 
were  carried  down  and  the  footing  under  the  corner  column  carry- 
ing the  least  dead  load  was  designed  for  the  dead  and  live  load. 
The  allowable  soil  pressure  was  multiplied  by  the  per  cent  of 
dead  load  brought  down  to  this  footing,  to  obtain  a  "soil  factor." 
The  soil  factor  was  divided  by  the  percentage  of  dead  load  brought 
down  for  each  column  and  thus  was  obtained  a  new  allowable  soil 
pressure  for  each  column.  With  these  allowable  soil  pressures 
as  thus  determined  for  each  footing  the  footings  were  designed 
for  the  dead  and  live  load.  This  method  is  really  practically  the 
same  as  that  proposed  by  Mr.  Schneider,  except  that  the  column 
loaded  the  most  heavily  with  dead  load  is  the  critical  column, 
whereas  with  the  Schneider  method  the  column  loaded  the  most 
heavily  with  live  load  is  the  critical  column.  The  method  so 
long  used  by  the  author  is  preferred  by  him,  but  instead  of  the 
corner  column  carrying  the  least  dead  load  he  selects  that 
column  on  which  the  live  load  is  not  less  than  15  per  cent.  This 
will  be  sufficient  for  many  buildings.  When  there  is  much  ma- 
chinery in  a  building  and  the  building  is  occupied  by  large  num- 
bers of  employees  for  eight  hours  and  is  closed  for  sixteen  hours 


264  PRACTICAL   STRUCTURAL   DESIGN 

each  day,  that  column  is  selected  on  which  the  live  load  is  from 
20  to  25  per  cent,  to  allow  for  the  constant  machine  load.  The 
student  can  see  that  there  is  considerable  room  for  the  exercise 
of  judgment  in  the  matter,  provided  the  fact  is  recognized  that 
to  design  all  footings  for  the  sum  of  the  dead  and  live  loads  is 
wrong. 

The  load  on  walls  is  assumed  to  be  one  foot  long.    The  load  on 
one  lineal  foot  of  wall  is  divided  by  the  allowable  soil  pressure 
per  square  foot  and  the  width  I  of  the 
footing  found,    Fig.    170.     It    is   then 
stepped.    An  old  rule  was  to  draw  a  line 
r°*|         upward,  at  an  angle  of  60  degrees  from 
I  £      the  end  of   the   footing  to    the    lower 
^1  *     corner  of  the  wall  and  form  steps  so  the 
.'l'^"**     line  touched  the  inner  corners.     By  as- 
suming  a  safe  fiber  stress  for  the  masonry 
Fig.  170 —  anci  computing  the  offsets  as  projecting 

Stepped    Masonry   Footing    cantileyers  the  thickness  and  projections 

of  the  steps  may  be  computed.  The  following  formula  is  used 
generally  by  designers;  referring  to  Fig.  170: 

o  =  offset  in  inches, 

t  =  thickness  in  inches, 

p  =  allowable  soil  pressure  in  pounds  per  sq.  in., 

s  =  safe  unit  tensile  stress  in  the  material, 
=  30  Ibs.  per  sq.  in.  for  1-3-5  concrete, 
=  60  Ibs.  per  sq.  in.  for  1-2-4  concrete. 

Various  authorities  give  values  for  stone  and  for  brick  laid  in 
cement  mortar.  The  values  for  stone  cannot  be  used  unless  the 
stone  projects  less  than  one-half  its  length  beyond  the  step  above. 
This  is  to  provide  for  true  cantilever  action.  If  built  in  this  way 
s  =  80  to  130  Ibs.  per  sq.  in.  for  limestone,  the  same  for  sandstone 
and  180  Ibs.  per  sq.  in.  for  granite.  Hard-burned  brick  laid  up  in 
cement  mortar  in  good  bond  by  a  first-class  mason  is  considered 
to  be  good  for  40  Ibs.  per  sq.  in.  The  author  advises  the  use  of 
concrete. 


The  formula  is  derived  as  follows: 


P^  x  2  =  P, 
144  X  2      288 


COLUMNS  AND  STRUCTURES  265 

for  p  is  in  pounds  per  square  feet  and  o  is  in  inches,  so  it  is  neces- 
sary to  reduce  p  to  pounds  per  square  inch.  The  load  is  uni- 
formly distributed  along  the  cantilever  o  =  po,  and  the  force 

acts  through  the  center  of  gravity  =  5-     The  moment  of  resistance 

sbtz 
for  a  rectangular  section  =  -^->  but,  since  6  =  1  (the  unit  width) 

the  equation  becomes 

Mb  =  Mr  =  2!  =  f ' 
Dividing,  ^r  =  sf2. 

Multiplying,  po2  =  48st2. 

48s£2      3  x 

Dividing  o2   = =  — 

p  p 

Extracting  the  square  root,  o  = 

The  assumption  is  made  in  this  case  that  the  projection,  or 
offset,  is  a  cantilever  with  the  maxi-  w 

mum  moment  at  the  face  of  the  T-, 

support.     This  is  true  in  the  case  CTlc- — y  — ->| 

of  all   stepped  footings  for  walls.      r —— — '      ' 1 

When  the  wall   is  relatively  thin, 
the  projection  of  the  footing  being 
long,  the  maximum  moment  is  as- 
sumed to  be  in  the  center  of  the  Fij?-  171  — 
footing  under  the  center  of  the  wall. 

The  formula  for  footings  such  as  grillage  beams  and  reinforced 
concrete,  referring  to  Fig.  171,  is 

"  ~4~* 

The  load  is  assumed  to  be  uniformly  distributed  over  the 
pressed  area.  The  formula  may  be  derived  in  two  ways. 

First.  Assume  one-half  the  load  to  be  carried  on  one  project- 
ing end  having  a  length  =  — ^ —  =  y 

_  W     y  _Wy 

Second.  Assume  the  slab  to  be  a  freely  supported  beam  with  a 
span  length  =  L  and  loaded  in  the  middle  with  a  load  W  occupy- 
ing a  width  a. 


266  PRACTICAL  STRUCTURAL   DESIGN 

M  -  WL  _  Wa      2WL      Wa      W^L  ~  a)  -E2 

4    "    8    =       8  8    =  8          =    4  ' 

since  (L  -  a)  =  2y. 

Not  all  authorities  agree  that  the  above  formula  for  the  design 
of  footings  is  correct,  the  contention  being  that  the  maximum 
moment  is  at  the  face  of  the  wall,  and  not  under  the  center.  The 
following  formula  is  used  for  this  assumption  : 


h  -  = 

2y  +  a  A  2      2(2y  +  a) 

To  design  a  wall  footing  let  w  =  weight  per  lineal  foot  of  wall 
and  divide  this  by  the  safe  soil  load  per  square  foot.  This  will 
give  the  width  L  of  the  footing.  The  bending  moment  being  ob- 

tained the  total  thickness 
will  be  i  and  the  total  off- 
set is  o,  in  the  formula  for 
stepped  footings.  This  will 
tne  thickness  at  the 


Y  -  *    face  of  the  wall  and  it  may 

be  stepped  off  by  dividing 
Fig.  172-  Design  of  Stepped  Footings       .,.  ^  any  number  Qf  ^ 

each  with  a  thickness,  t,  and  solving  for  o  by  the  formula  for 
each  step. 

Such  footings  are  seldom  so  designed.  The  usual  way  is  to 
ascertain  the  width  and  then  from  the  edge  of  the  bottom  of  the 
wall  draw  a  line  at  an  angle  of  45  degrees,  or  60  degrees,  with 
the  horizontal  until  the  horizontal  distance  separating  the  lines 
is  equal  to  the  spread  of  the  footing.  Steps  are  then  drawn  to 
touch  this  line,  as  shown  in  Fig.  172. 

The  real  use  made  of  the  formulas  for  bending  moment  on 
footings  is  to  determine  the  size  of  I  beams  to  use  in  a  grillage 
foundation,  or  the  thickness  and  reinforcement  for  a  reinforced 
concrete  footing.  The  design  of  grillage  footings  is  given  in  the 
steel  handbooks.  A  reinforced  concrete  footing  is  designed  as  a 
slab  one  foot  wide,  the  thickness  being  fixed  both  by  shear  and 
bending. 

Column  footings  differ  from  wall  footings  in  being  square  or 
rectangular.  If  the  footing  is  square  the  area  is  found  by  dividing 
the  total  load  by  the  allowable  soil  pressure.  The  square  root  of 
this  area  gives  the  length  of  each  side.  If  the  side  of  the  column 


COLUMNS   AND   STRUCTURES 


267 


is  longer  than  the  end  and  it  is  desired  to  proportion  the  ends  and 
sides  of  the  footing  in  the  same  ratio  the  following  procedure  is 
adopted. 

Let  L  =  the  long  side, 
6  =  the  short  side, 
A  =  the  area  of  the  footing  in  square  feet. 

Then  -r-  =  o,  which  is  the  ratio  of  the  length  and  breadth  of  the 

column  area  and  of  the  area  of  the  footing,  that  is, 
L  =  ab, 
A  =  bL  =  b  +  ab  =  ab2, 

'.-vf 

Having  obtained  the  length  and  breadth  of  the  footing  it  is 
assumed  that  there  are  four  cantilever  beams  projecting  from 


Fig.  173  —  The  Design  of  Column  Footings 

the  column.  Each  has  a  width  at  one  end  equal  to  the  width  of 
the  column  base  and  a  width  at  the  other  end  equal  to  the  length 
of  the  side.  The  beams  are  thus  wedge  shaped  with  the  maxi- 
mum moment  at  the  narrow  end.  Each  beam  carries  one-fourth 
of  the  total  load.  For  convenience  the  beam  may  be  considered 
to  be  divided  into  four  strips,  as  shown  in  Fig.  173. 

Referring  to  the  figure  and  assuming  that  in  this  case  one- 
fourth  the  load  is  carried  on  each  beam,  then  the  beam  ajfe  carries 

W 

— •     Each  strip  carries  a  part  of  this  load  in  proportion  to  the 

area  of  the  strip.  The  strips  may  be  of  equal  width,  as  shown, 
or  they  may  be  varied  in  order  to  have  equal  loads  on  the  strips. 


268  PRACTICAL  STRUCTURAL   DESIGN 

Each  load  is  assumed  to  be  concentrated  at  the  center  of  gravity 
of  each  strip  and  the  projection  y  is  assumed  to  be  the  length  of 
a  cantilever  beam.  The  moment  across  the  beam  on  the  lines 
bi,  ch,  dg,  and  ef,  can  be  readily  ascertained  and  the  shears  can 
also  be  found  on  these  lines.  In  this  example  we  are  assuming  a 
reinforced  concrete  beam.  The  thickness  of  the  footing  on  each 
line  can  be  found  in  the  usual  way  for  the  design  of  a  concrete 
beam,  or  slab,  both  for  bending  stress  and  for  shear.  The  steel 
may  be  proportioned  and  the  logical  method  for  arranging  the 
steel  would  be  to  have  it  in  four  layers,  two  normal  to  the  sides 
of  the  slab  and  two  diagonal,  as  indicated  by  the  dotted  lines  in 
the  lower  part  of  Fig.  173. 

Another  method,  used  by  the  author,  is  to  take  the  expres- 
sion M  =  -~f  used  for  wall  footings,  and  assume  that  the  foot- 
ing being  square  and  there  being  eight  projections  instead  of  two, 
the  expression  should  be  M  =  -r^>  and  use  this  moment  to  design 

a  reinforced  concrete  beam  having  a  width  equal  to  the  column 
base  on  top  of  the  footing.  The  beams  are  considered  as  being 
so  arranged  that  two  are  normal  to  the  sides  of  the  footing  and 
two  are  diagonal.  They  are  designed  as  reinforced  concrete  beams 
and  as  merged  so  that  while  each  layer  of  steel  carries  the  tension 
for  the  beam  it  represents,  the  concrete  is  stressed  in  compression 
from  all  directions,  which  makes  it  safe  and  increases  its  resist- 
ance to  shear. 

Reinforced  concrete  footings  may  be  stepped  or  sloped  on  top. 
If  sloped  the  forms  must  be  well  anchored  down,  for  the  con- 
crete will  have  a  tendency  to  cause  them  to  float.  The  author 
obtained  the  best  results  with  concrete  footings  by  stepping  them. 
The  steps  are  formed  by  frames  of  boards.  The  first  step  is  cast 
to  the  proper  level  and  the  frame  for  the  next  step  placed  on  it, 
when  it  becomes  firm  enough  to  carry  the  weight  of  the  next  step 
without  bulging  up  around  the  edges  of  the  form.  If  the  con- 
crete is  mixed  to  the  proper  consistency  there  will  never  be  any 
trouble  with  this  bulging  and  the  steps  can  be  poured  quickly. 
The  proper  consistency  for  concrete  is  that  of  a  soft  tooth  paste. 
It  should  never  be  thin  enough  to  pour  into  a  form.  For  rein- 
forced concrete  it  should  be  thin  enough  to  flow  very  sluggishly 
so  it  will  surround  the  reinforcement,  but  it  should  never  be  so 


COLUMNS   AND   STRUCTURES  269 

thin  that  the  aggregates  have  a  tendency  to  separate.  Get  in 
touch  with  the  Portland  Cement  Association,  Chicago,  111.,  when 
reliable  information  on  concrete  is  required.  The  Association 
has  a  well  equipped  laboratory  where  all  questions  affecting  the 
use  of  portland  cement  and  the  manufacture  and  use  of  concrete 
are  investigated. 

Column  Brackets  and  Bases 

Brackets  on  steel  columns  for  carrying  beams  and  girders  are 
rivetted  to  the  columns.  For  light  loads  they  are  simple  shelf 
angles.  For  heavy  loads  they  consist  of  plates  stiffened  with 
angles.  Details  are  given  in  the  steel  handbooks. 

Post  caps  for  wooden  columns  are  illustrated  in  a  book  entitled 
"Heavy  Timber  Mill  Construction  Buildings"  distributed  free 
of  cost  by  the  National  Lumber  Manufacturers  Bureau,  Chicago, 
111.  Mill  construction  has  its  place  and  every  architectural  de- 
signer should  have  a  copy  of  the  book.  Some  of  the  statements 
therein  should  be  modified  and  the  student  is  advised  to  obtain 
from  the  Portland  Cement  Association,  Chicago,  111.,  a  bulletin 
entitled  "Why  Build  Fireproof,"  written  by  the  author,  and 
thus  obtain  a  glance  at  both  sides  of  the  question.  The  book  on 
mill  construction  illustrates  cast  iron  and  steel  post  caps  for 
carrying  beams  and  girders.  The  girders  do  not  rest  on  top  of 
the  posts,  for  the  carrying  power  of  the  posts  would  be  reduced 
thereby.  It  was  formerly  the  custom  to  use  wooden  bolsters, 
on  which  to  rest  the  girders  and  beams,  in  order  to  shorten  the 
span.  The  side  bearing  strength  of  wood  is  much  lower  than  the 
strength  with  the  fibers.  When  the  direct  load  comes  down 
the  post  with  a  bearing  stress,  say,  of  1100  Ibs.  per  square  inch, 
and  rests  on  the  side  of  a  bolster  the  bearing  stress  is  reduced  at 
once  to  235  pounds  per  square  inch,  or  more,  depending  upon 
the  wood.  Bolsters  to-day  are  used  only  under  roof  girders.  They 
are  better  in  case  of  fire  than  cast  iron  or  steel  but  so  greatly 
reduce  the  carrying  power  of  the  posts  that  they  are  not  eco- 
nomical. Post  caps  come  in  a  number  of  different  shapes.  To  in- 
vestigate the  strength  of  a  post  cap  obtain  the  moment  of  inertia 
of  the  cross  section  of  the  carrying  portion.  Then  find  the  fiber 
stress  by  the  method  shown  on  page  67.  The  bending  moment  is 
the  moment  caused  by  the  reaction  from  the  beam  acting  at  half 
the  length  of  the  projection  from  the  face  of  the  column. 


270 


PRACTICAL   STRUCTURAL   DESIGN 


Cast  iron  brackets  are  cast  on  the  side  of  cast  iron  columns. 
There  is  really  no  rational  method  for  designing  them.  They 
should  be  at  least  as  thick  as  the  shell  of  the  column.  For  light 
loads  one  bracket  is  used  under  the  shelf  and  for  wide  beams  two, 
or  more,  brackets  are  used  to  avoid  eccentricity  in  the  loading. 
The  depth  of  the  brackets  at  the  face 
of  the  column  must  be  enough  to  pre- 
vent shearing.  There  is  of  course  a 
bending  moment  created  by  the  reac- 
tion from  the  beam.  This  bending 
moment  divided  by  the  depth  of  the 
bracket  gives  the  tension  in  the  shelf 
where  it  is  attached  to  the  shell  of 
the  column.  This  tension  must  be 
divided  by  the  length  of  the  edges  of 
the  shelf  and  multiplied  by  the  thick- 
ness of  the  metal  in  the  shell  to  de- 
termine the  shearing  stress  which  acts 
to  tear  the  shelf  away  from  the  shell. 
This  shearing  stress  divided  by  the 
safe  allowable  shear  in  the  metal 


Fig.  174—  Cast  Iron,  or  Steel, 
Ribbed  Column  Base 


must  be  less  than  the  tension,  or  the  thickness  of  the  shelf  must 
be  increased.  Such  computations  are  merely  checks  but  should 
not  be  neglected. 

In  Fig.  174  is  illustrated  a  ribbed  cast  iron  base  cap  for  a 
column.  No  known  rational  method  exists  for  determining  the 
stresses,  so  these  bases  are  made  according  to  empirical  rules. 
The  thickness  of  all  the  parts  should  be  not  less  than  the  thick- 
ness of  the  shell  of  the  column.  All  parts  should  have  the  same 
thickness  to  avoid  danger  of  casting  cracks.  A  small  fillet  should 
be  used  in  every  angle. 

The  projection  at  the  top  should  be  not  less  than  three  inches 
wide,  so  bolts  can  be  used  with  plenty  of  clearance  for  the  heads. 
When  the  bottom  projection  P  is  greater  than  six  inches,  ribs  should 
be  used.  The  height  H  should  never  be  less  than  the  projecting  P 
and  the  diameter  of  the  base  under  the  column  should  be  equal  to 
that  of  the  column.  The  number  of  ribs  should  never  be  less  than 
eight,  and  an  empirical  rule  for  fixing  the  number  of  ribs  is  that 
the  space  between  ribs  at  the  circumference  of  the  column  should 
never  be  greater  than  twice  the  thickness  of  the  shell. 


COLUMNS  AND   STRUCTURES  271 

Cast  steel  bases  are  better  than  cast  iron  bases  and  built-up 
steel  bases  are  a  great  deal  better  than  cast  steel  bases.  Cast 
steel  bases  are  designed  in  the  same  way  as  cast  iron  bases.  Built- 
up  steel  bases  are  illustrated  in  the  steel  handbooks. 

When  the  projection  P  of  a  cast  iron,  or  cast  steel,  base  is  not 
greater  than  six  inches  a  plate  may  often  be  used  to  advantage. 
The  formula  to  use  in  the  design  of  such  a  plate  is  given  on  page 
152,  where  it  is  used  to  design  a  washer  under  the  head  of  a  bolt. 
Mention  is  there  made  of  ribs  acting  as  cantilevers  and  if  the 
student  wishes  to  attempt  to  design  bearing  plates  and  bases 
of  cast  iron  according  to  formulas,  he  is  advised  to  procure  from 
the  Engineering  Experiment  Station,  University  of  Illinois,  Ur- 
bana,  111.,  a  copy  of  the  bulletin  on  the  design  of  cast  iron  column 
bases  and  bearing  plates.  He  is  advised  also  to  procure  a  copy 
of  Bulletin  No.  67  on  the  design  of  reinforced  concrete  footings. 

The  size  of  a  column  is  fixed  by  the  compressive  strength  of 
the  material.  The  column  rests  on  concrete,  stone  or  brick  foot- 
ings, which  have  a  lower  strength  in  compression,  so  it  becomes 
necessary  to  enlarge  the  lower  end  in  order  not  to  overstress  the 
masonry.  It  is  most  convenient,  in  the  majority  of  cases,  to  set 
the  spread  base  before  erecting  the  column,  so  bases  are  made 
to  which  the  columns  are  bolted.  The  area  of  the  bottom  of  the 
base  is  obtained  by  dividing  the  load  by  the  bearing  strength  of 
the  masonry. 

Eccentric  Loads  on  Footings 

In  Fig.  175  is  illustrated  a  common  case  of  eccentric  loading 
on  the  footing  of  a  wall.  The  formula  to  use  is  given  on  pages 
100  and  101.  The  direct  load  divided  by  the  area  of  the  footing 
gives  the  pressure  per  square  foot  on  the  soil.  A  vertical  line  is 
drawn  through  the  center  of  gravity  of  the  footing  and  a  vertical 
line  is  drawn  through  the  center  of  gravity  of  the  wall.  .  The 
horizontal  distance  e  is  the  moment  arm.  Multiply  the  load  in 
pounds  by  the  moment  arm,  e,  in  feet  and  use  the  bending  mo- 
ment in  the  formula.  In  the  formula  h  is  used  as  the  depth  of 
the  member.  In  the  case  of  the  footing  h  is  the  width,  I  shown 
in  the  figure.  The  resulting  fiber  stress  cannot  exceed  the  safe 
allowable  bearing  pressure  on  the  soil.  A  positive  (+)  result 
indicates  compression  and  a  negative  (— )  sign  indicates  uplift. 
The  two  shaded  diagrams  illustrate  the  action.  The  upper  one 


272 


PRACTICAL   STRUCTURAL   DESIGN 


shows  the  load  per  square  foot  over  the  footing,  provided  the 
load  is  applied  through  the  center  of  gravity.  The  lower  diagram 
shows  the  effect  of  the  bending  moment  to  increase  the  compres- 
sion on  one  side  and  lessen  it  on  the  other. 

If  the  sum  of  the  two  does  not  exceed  the  safe  allowable  pres- 
sure at  one  toe  and  there  is  no  uplift  (tension)  on  the  other  toe 
the  footing  will  be  safe.  The  rule  to  as- 
sure safety  is  known  as  the  "Middle-third 
rule"  in  which  the  resultant  pressure  to 
prevent  overturning  must  be  kept  within 
the  middle  third  of  the  base.  The  middle- 
third  rule  has  been  considerably  over- 
worked. What  it  really  amounts  to  is  a 
statement  that  if  the  resultant  of  all  pres- 
sures brought  to  bear  on  a  footing  base  is 
kept  within  the  middle  third  the  average 
stress  will  not  exceed  one-half  the  maxi- 
mum and  there  will  be  no  tension. 

The  condition  shown  in  Fig.  175  is  not 
always  possible  to  avoid,  for  foundations 
must  be  kept  within  lot  lines.  The  remedy 


Fig.  175 — Eccentric  Load 
on  Wall  Footing 


apparently  is  to  so  construct  the  footing  that  a  line  may  be  drawn 
at  an  angle  of  thirty  degrees,  with  the  vertical,  from  the  edge  of 
the  wall  to  the  lower  edge  of  the  footing  and  keep  within  the 
footing.  When  the  load  brought  down  by  the  wall  reaches  the 
footing  it  will  spread  out  and  thus  the  center  of  effort  of  the  load 
will  not  be  directly  under  the  center  of  the  wall,  but  will  be  some- 
what nearer  the  center  of  gravity  of  the  footing.  This  may  be 
the  case  if  the  wall  is  not  in  excavation.  If  it  is  in  excavation 
the  load  is  no  doubt  partly  distributed  to  the  earth  on  the  out- 
side of  the  wall,  so  the  center  of  effort  of  the  load  is  actually 
under  the  center  of  the  wall,  provided  the  load  passes  as  readily 
through  the  masonry  as  it  does  through  the  earth  on  the  side. 

The  footing  may  be  of  solid  concrete,  with  a  depth  fixed  by  the 
sixty  degree  line,  so  that  it  will  not  distort  under  load.  If  this 
is  the  case,  then  as  soon  as  the  earth  under  the  heavily  pressed 
edge  gives  way  the  entire  bottom  of  the  footing  will  come  into 
bearing  and  relieve  the  stress  on  the  soil.  The  same  effect  should 
possibly  be  secured  by  using  a  lighter  footing  of  concrete  heavily 
reinforced  so  it  cannot  bend.  Something  also  may  be  gained 


COLUMNS   AND   STRUCTURES  273 

by  having  the  inner  toe  deeper  than  the  outer  instead  of  keeping 
the  base  level. 

The  best  remedy  is  to  drive  piling  to  help  the  earth  carry  any 
excessive  load.  In  several  of  the  specifications  mentioned,  rules 
are  given  for  the  use  of  piles.  A  pile  acts  by  the  bearing  of  the 
lower  end  of  the  pile  on  the  soil  into  which  it  is  driven,  plus  the 
friction  of  the  pressed  soil  on  the  surface  of  the  pile.  The  mini- 
mum distance  center  to  center  of  piles  should  not  be  less  than 
three  feet,  except  under  unusual  conditions.  Driving  piles  too 
closely  together  often  results  in  an  actual  lessening  of  the  carry- 
ing capacity. 

To  find  the  center  of  gravity  of  a  stepped  footing  such  as  that 
shown  in  Fig.  175  multiply  the  area  of  each  strip  by  half  the  dis- 
tance from  the  outer  edge.  Add  the  results  and  divide  by  the 
total  area.  This  gives  the  horizontal  distance  to  the  center  of 
gravity.  The  distance  from  the  bottom  is  found  similarly  by 
dividing  the  area  into  strips  by  vertical  lines  and  multiplying 
each  area  by  half  the  depth.  It  is  the  method  of  moments,  al- 
ready explained  for  irregular  sections  and  for  bending  moments 
on  rivets.  The  center  of  gravity  is  not  essential  in  the  footing 
problem,  but  it  is  essential  to  obtain  the  horizontal  distance  to 
a  vertical  line  through  the  center  of  gravity. 

Eccentric  Loads  on  Column  Base 

A  column  exposed  to  the  force  of  wind  acting  to  push  it  to  one 
side  will  put  an  eccentric  load  on  the  base.  There  are  two  cases. 

I.  The  column  hinged  to  the  base.     The  horizontal  force  in 
this  case  acts  at  the  base  of  the  column.    The  vertical  force  acts 
-vertically  through  the  center  of  the  base  and  the  resultant  of  the 
horizontal  and  vertical  forces  should  be  kept  within  the  middle 
third  if  possible. 

II.  The  column  is  fastened  to  the  base  sufficiently  to  make  the 
base  practically  a  part  of  the  column,  or  at  least  develop  a  bend- 
ing moment  =  H  x  «'  m  which 

H  =  horizontal  thrust, 
I  =  length  of  column. 

In  Case  I  the  horizontal  force  acts  at  the  top  of  the  footing, 
as  shown  in  Fig.  176.  In  Case  II  it  acts  halfway  up  on  the 
column.  It  is  important  to  remember  these  distinctions.  The 


274 


PRACTICAL   STRUCTURAL   DESIGN 


(0 


distance  x  on  the  bottom  of  the  footing  should  in  all  cases  be  not 
more  than  -  the  base,  to  keep  the  average  stress  within  one-half 

the  maximum  and  insure  that  there  is  no  tension,  or  uplift,  on 
the  other  side.  This  "middle  third  theory"  is  merely  a  state- 
ment. Provided  the  maximum  soil  pressure  is  not  greater  than 

the  allowable  safe 
pressure  the  re- 
sultant  can  be 
outside  the  middle 
third.  The  state- 
ment is  frequently 
made  that  when 
the  resultant 
passes  beyond  the 
]  *-H  \  middle  third  the 

Fig.  1 76  —  Eccentric  Load  on  Column  Bases  structure  is  in 

danger    of    being 

overturned.  The  fact  is  that  whenever  the  resultant  of  a  hori- 
zontal and  a  vertical  force  passes  through  a  footing  at  any  point 
off  center  the  overturning  tendency  is  present.  To  be  safe  it  is 
necessary  to  see  that  no  undue  load  is  placed  on  the  soil.  The 
"middle  third"  theory  is  a  safe  one  to  follow,  but  it  should  not 
be  followed  blindly. 
Let  H  =  horizontal  force, 

h  =  distance  at  which  H  acts  above  bottom  of  base, 
W  =  vertical  load  on  footing, 
Hh 


then  x  = 


W 


Let  6  =  length  of  base  (that  is,  the  dimension  at  right  angle  to 
the  force  H,  the  dimension  B  being  in  the  direction 
of  the  force  H). 

p  =  pressure  on  soil  hi  pounds  per  sq.  ft.,  all  the  dimensions 
being  expressed  in  feet  and  the  weight  and  wind  force 
in  pounds; 
then,  when  the  resultant  falls  within  the  middle  third, 


When  the  resultant  falls  beyond  the  middle  third 

2w 


COLUMNS   AND  STRUCTURES  275 

The  student  should  study  the  formation  of  the  last  three 
formulas.  The  first  one  means  that  a  force  H,  acting  with  an 
arm  h,  tends  to  overturn  a  body  having  a  weight  W.  It  is, 
therefore,  necessary  to  find  the  length  of  an  arm  x,  through  which 
W  acts  to  resist  the  overturning  moment  Hh. 

In  the  second  the  distance  x  cannot  be  greater  than  one-sixth 
of  B.  The  total  weight  is  distributed  over  an  area  Bb.  With 
these  hints  the  student  should  attempt  to  construct  the  formulas 
as  an  exercise. 

Attaching  Column  Bases  to  Footings 

Column  bases  are  attached  to  footings  by  bolts.  A  horizontal 
force,  such  as  wind,  develops  a  bending  moment  in  the  column 
where  it  is  attached  to  the  footing.  Referring 
to  Fig.  177,  divide  the  bending  moment  by 
the  distance  x  between  the  center  lines  of  the 
bolts  in  the  direction  of  the  force.  This  gives 
the  pull  on  the  bolts.  The  distance  is  mea- 
sured between  centers  of  bolts  instead  of  from 
the  leeward  edge  of  the  plate  to  the  center 
line  of  the  windward  bolts,  to  avoid  bending 
the  edge  of  the  plate. 

The  pull  in  the  bolts  is  divided  by  the  allow- 
able tensile  fiber  stress,  to  obtain  the  required  bolt  area.  Divid- 
ing this  by  two  gives  the  area  of  one  bolt,  and  the  circumference 
is  readily  found  when  the  diameter  is  known.  Dividing  the  cir- 
cumference in  inches  by  50  Ibs.  gives  the  total  bond  resistance  per 
lineal  inch  of  bolt.  Dividing  the  uplift  on  one  bolt  by  this 
amount,  the  length  of  bolt  is  obtained. 

The  area  of  footing  is  determined  by  the  bearing  value  of  the 
soil.  The  depth  must  be  great  enough  to  make  the  footing  of 
the  weight  required  and  also  furnish  area  for  embedment  of 
bolts.  The  weight  of  the  footing  is  fixed  by  the  requirement  that 
it  be  heavy  enough  to  anchor  the  structure,  or  so  much  of  the 
structure  as  may  bfe  carried  by  the  column  which  rests  on  the 
footing. 

The  horizontal  force  acting  on  the  column  is  applied  to  the 
column  at  the  proper  height.  The  force  multiplied  by  the  height 
exerts  an  overturning  moment.  Dividing  this  moment  by  the 
width  of  the  building  the  weight  of  the  foundation  is  obtained. 


276  PRACTICAL   STRUCTURAL   DESIGN 

This  is  illustrated  in  Fig.  178.  The  direct  weight  W  is  made 
up  of  the  weight  carried  to  the  foundation  by  the  column,  plus 
the  overturning  moment  exerted  by  the  force  H  acting  on  the 
windward  column.  The  force  H,  acting  on  the  leeward  column 
is  an  eccentric  load  on  the  footing,  as  already  described.  The 

student  must  remember 
that  each  column  carries 
part  of  the  total  horizontal 
thrust. 

Owing  to  the  uncertainty 
of  just  where  the  force  will 
act  because  the  attachment 


may  not  be  rigid,  there  is 

'  ,_0     _       ,  ,.  some  uncertainty  as  to  the 

Pig.  1/8  —  foundations  under  Columns  , 

exact  amount  of  force  ex- 
erted on  footings.  If  the  columns  are  fixed  (hinged)  in  such  a 
way  that  they  bend  at  the  top  of  the  footing,  the  only  force 
exerted  by  wind  on  a  footing  will  be  that  transmitted  by  the 
windward  column  as  a  direct  vertical  load.  If  the  column  is 
rigidly  attached  then  the  leeward  column  adds  an  eccentric  load. 

If  the  columns  are  rigidly  attached  the  force  H  acts  at  a  height 
equal  to  one-half  the  column  length.  If  there  is  a  knee  brace 
some  men  assume  the  length  I  to  be  the  distance  from  the  base 
to  the  lower  end  of  the  knee  brace.  To  be  safe  it  is  best  to  con- 
sider the  column  length  to  be  measured  to  the  bottom  chord  of 
the  truss. 

If  the  columns  are  poorly  connected,  or  hinged,  to  the  footing, 
the  total  wind  force  on  the  windward  column  is  assumed  to  act 
at  a  height  equal  to  the  full  column  length,  plus  one-half  the 
height  of  the  roof  truss. 

For  taller  buildings  than  here  illustrated  the  total  force  of  the 
wind  is  assumed  to  act  at  half  the  height  of  the  building,  or  if 
only  the  upper  portion  of  the  building  is  exposed,  at  half  the 
height  of  the  exposed  portion,  measured  from  the  ground,  the 
full  amount  of  wind  being  figured  only  on  the  exposed  portion. 
This  moment  divided  by  the  width  of  the  building  gives  the 
weight  which  must  be  opposed  to  resist  overturning.  Determine 
the  direct  weight  coming  on  the  footing  and  if  it  is  not  enough 
increase  the  depth  of  the  footing  so  it  will  have  enough  weight. 


COLUMNS  AND   STRUCTURES  277 

Cantilever  Footing 

The  subject  of  cantilever  footings  is  very  simple,  although  a 
number  of  students  seem  to  find  it  difficult.  In  Fig.  179  the 
column  on  the  right  is  against  a  wall,  or  property  line,  and  the 
footing  must  be  kept  within  the  limits  of  the  property.  The 
first  line  of  ulterior  columns 
must,  therefore,  help  out  the 
wall  columns.  First  find  the 
proper  size  of  footing  under 
each  line  of  columns.  The 


outside  footing  is  arranged  so    ;  | 

the  outer  face  is  even  with      .  i       I  •  i/,,     "  1 

//,.  ,,      //.  ////  I  \w///",   i 


the  lot  line.    The  load  P,  com-    l^f    \#'""  \  r*-\«*  j 


ing  down  the  wall  column,  acts         1       l>  x         I 

at  the  center  of  the  column         Fig  179  _  Cantilever  Footings 
and  the  distance  to  the  center 

of  the  footing  gives  the  moment  due  to  eccentric  loading,  thus 
M  =  Pa. 

Divide  this  by  the  load  in  the  interior  column  to  obtain  the 
eccentricity  6.     The  footing  under  the  interior  column  is  then 
so  located  that  the  center  of  the  footing  will  be  distant  b  feet 
from  the  center  line  of  the  column.    Thus, 
.       Pa 

b  =  pT- 

To  transfer  the  loads  from  the  columns  to  the  centers  of  the 
respective  footings  the  columns  rest  on  a  girder  having  a  resist- 
ing moment  Mr  =  Mt,  =  Pa  =  P'b.  The  girder  must  also  be  de- 
signed for  deflection,  as  a  cantilever  at  each  end,  and  for  shear 
which  is  usually  very  high. 

"  Foundations  of  Bridges  and  Buildings,"  by  Jacoby  and  Davis, 
price  $7.50,  is  the  best  book  on  the  subject  with  which  the  writer 
is  acquainted.  The  student  is  advised  to  consult  it  if  he  needs 
more  information  on  this  important  subject. 

Stresses  in  Towers 

A  tank  tower,  or  any  braced  tower  or  pole,  may  be  designed 
as  a  vertical  cantilever  beam  or  truss. 

The  vertical  load,  consisting  of  the  weight  of  the  water,  tank, 
and  framework,  is  transmitted  directly  to  the  foundations  through 


278 


PRACTICAL   SPRUCTURAL   DESIGN 


the  columns,  each  carrying  its  proportionate  share  of  the  load. 
This  vertical  load  is  not  considered  in  the  graphical  stress  dia- 
gram used  to  determine  the  force  due  to  wind,  so  must  be  added 
to  the  wind  forces  after  the  stress  diagram  is  constructed  and 
scaled.  The  struts  and  ties  carry  no  part  of  the  tank  and  water 
load,  being  used  to  take  care  of  the  wind  load  and  to  divide  the 
column  into  intermediate  lengths  so  the  columns  may  be  as 


WIND 

Stresses  in 
Struts  and Anc  or/ 
Botts  a 


5000  b 


A  ,  . 

k 33-0    ->J 

Tower   Diagram. 

Fig.  180  —  Stresses  in  Water  Tower 

small  as  possible,  each  column  being  considered  as  having  a  length 
equal  to  the  height  of  one  bay. 

The  plan  of  the  base  in  Fig.  180  shows  that  the  maximum 
stresses  in  the  rods  and  struts  occur  when  the  wind  is  blowing 
against  the  side  of  the  tower.  The  maximum  stresses  in  columns 
occur  when  the  wind  is  blowing  diagonally  across  the  tower  and 
are  0.707  of  the  amounts  obtained  from  the  diagram.  The  wind 
force  is  not  the  result  of  a  constant  load,  so  in  a  number  of  specifi- 
cations only  80  per  cent  of  the  maximum  wind  force  is  considered. 

The  wind  acts  at  the  top  through  the  center  of  gravity  of  the 
filled  tank,  and  at  the  joints  on  the  side  of  the  tower.  In  the 


COLUMNS  AND   STRUCTURES  279 

tower  diagram  dotted  lines  are  shown,  transmitting  the  wind 
from  the  tank  to  the  frame.  In  the  stress  diagram  the  uplift  is 
measured  from  /  to  I.  The  tower  loads  are  all  positive  (compres- 
sion). In  the  tower  diagram  dotted  counters  are  shown.  These 
are  not  stressed  with  the  wind  coming  from  the  left,  but  are 
stressed  with  the  wind  coming  from  the  right.  Each  bay,  there- 
fore, is  braced  with  cross  braces  designed  both  for  tension  and 
compression.  The  stress  diagram  shown  is  only  one  method  for 
drawing  such  diagram. 

Wind  acts  against  the  side  of  a  square  or  rectangular  building 
over  the  whole  area.  Against  an  octagonal  structure  the  width 
is  measured  on  the  longest  possible  diagonal,  and  the  area  of  the 
height  multiplied  by  this  diagonal  is  multiplied  by  0.707,  for 
some  of  the  force  of  the  wind  is  lost  against  the  sloping  faces. 
Against  a  cylindrical  structure  more  of  the  force  is  lost  on  the 
curved  surface,  so  the  diameter  is  multiplied  by  the  height  and 
the  product  by  0.667.  In  old  text  books  the  statement  is  fre- 
quently met  with  that  the  pressure  of  the  wind  against  a  cylindri- 
cal surface  is  one-half  that  against  a  square  having  a  side  equal 
to  the  diameter  of  the  cylinder.  Exact  mathematical  analysis 
shows  it  to  be  two-thirds  instead  of  one-half. 

The  design  of  foundations  under  water  towers  is  similar  to 
the  design  of  foundations  under  the  columns  of  buildings.  The 
weight  of  the  water  and  tank  carried  down  by  each  column, 
added  to  the  weight  of  the  foundation,  must  be  sufficient  to  oppose 
the  overturning  effect  of  the  wind,  the  uplift.  It  is  necessary  to 
make  a  stress  diagram,  assuming  the  tank  to  be  empty.  It  can 
be  assumed  that  the  columns  are  rigidly  braced  so  there  will  be 
no  bending  to  set  up  an  eccentric  load  in  the  foundation  blocks. 
The  bolts  anchoring  the  column  footings  to  the  columns  must  be 
strong  enough  to  lift  the  weight  of  the  blocks. 

The  Design  of  Chimneys 

A  chimney  is  subjected  to  direct  stress  and  also  to  a  bending 
moment  caused  by  wind.  The  resultant  stress  on  the  leeward 
edge  must  not  exceed  the  safe  allowable  stress. 

>  The  formula  is  /  =  -j-  ±  -j-> 

in  which  W  =  weight  of  stack  above  section  considered, 
A  =  area  of  ring  in  square  inches, 


280  PRACTICAL   STRUCTURAL   DESIGN 

M  =  bending  moment  in  inch  pounds, 
c  =  distance  from  resultant  through  section  to  lee- 
ward edge,  in  inches, 
7  =  moment  of  inertia. 

The  quantity  c  is  of  importance  because  the  chimney  shaft  is 
hollow.  With  a  solid  symmetrical  section  the  distance  to  the 
most  stressed  fiber  is  measured  from  the  center  of  area,  that  is, 
the  center  of  gravity.  In  a  hollow  ring  or  square  the  pressure 
varying  uniformly  from  zero  on  the  windward  edge  to  a  maxi- 
mum on  the  leeward  edge  is  not  an  average  at  the  neutral  axis 
of  the  section,  but  is  an  average  at  a  point  a  trifle  beyond.  The 
distance  from  the  center  of  area  to  the  center  of  pressure  will 
be  termed  q. 

Let  D  =  outer  diameter  of  a  round  hollow  shaft  or  outside  length 

of  a  square  hollow  shaft. 

d  =  inner  diameter  of  a  round  hollow  shaft  or  inner  length 
of  a  square  hollow  shaft. 

D2  +  d2 
then,  for  a  round  hollow  shaft  q  =  — 5-^ — > 

oLf 

and  for  a  square  hollow  shaft  q  =  — ^-^ — 

Approximately  values  of  q  are  f  for  square  and  |  for  round 
shafts.  r» 


The  chimney  for  analysis  is  divided  into  a  number  of  sections 
and  the  horizontal  area  found  at  each  section.  The  weight  of 
the  chimney  above  any  section  is  ascertained  and  divided  by  the 
area  of  the  ring  to  get  the  direct  pressure.  .  The  vertical  area  of 
the  chimney  above  the  section  is  obtained  and  multiplied  by  the 
factor  0.707  if  octagonal  and  0.667  if  cylindrical.  The  height  to 
the  center  of  this  area,  measured  from  the  section,  is  a  moment 
arm,  by  which  the  wind  moment  is  obtained.  Then  find  c  and 
apply  the  formula.  The  compressive  fiber  stress  should  not  ex- 
ceed the  maximum  allowed  for  the  material  and  for  good  brick 
in  cement  mortar  there  can  be  some  tension  on  the  windward 
side  not  exceeding  one-tenth  the  compressive  stress,  provided 
the  tension  is  not  more  than  one-fifteenth  of  the  allowable  com- 
pressive stress  in  the  material.  With  lime  mortar  tension  is  not 
permissible. 


COLUMNS   AND   STRUCTURES  281 

The  material  is  generally  brick  and  the  maximum  allowable 
compressive  stress  is  fixed  by  building  ordinances  or  in  the  speci- 
fications governing  the  design. 
For  a  rough  check  of  a  design 
Let  P  =  the  total  wind  pressure  in  pounds, 

h  =  height  in  feet  to  the  center  of  gravity  of  the  shaft, 
W  =  weight  of  shaft  above  section, 
D  =  outer  dimension, 

WD 
then,  for  round  shafts  hP  =  — -r- >     (Approx.) 

WD 
and,  for  square  shafts  hP  =  —= —     (Approx.) 

o 

Having  tested  the  shaft  at  a  number  of  joints,  approximately 
twenty-five  feet  apart,  redesigning  the  walls  if  found  to  be  un- 
stable, or  if  the  maximum  pressure  exceeds  the  safe  allowable 
pressure  the  final  test  is  made  at  the  base,  on  top  of  the  founda- 
tion. 

The  spread  base  must  be  designed  to  carry  the  shaft  without 
exceeding  the  safe  allowable  soil  pressure.  It  may  be  square, 
octagonal,  or  round.  It  is  subjected  to  a  direct  load,  which  is  the 
weight  of  the  shaft.  The  weight  of  the  lining  is  neglected  for 
brick  and  concrete  chimneys.  It  is  subjected  to  an  eccentric 
load  due  to  the  moment  caused  by  the  wind.  Then, 

W    -M 
p'A±2f 

in  which  p  =  soil  pressure  in  pounds  per  square  foot, 
M  =  moment  in  foot  pounds, 

I  =  moment  of  inertia  in  square  feet, 

and  the  distance  to  the  most  stressed  fiber  is  one-half  the  length, 
or  width,  of  the  footing. 

.      .       ,    ,,      W  (D4  -  d4) 
/,  for  hollow  circular  shaft  =  — Vsj - 

D4  —  d* 
I,  for  hollow  square  shaft  =  — -~ — 

D4 

7,  for  square  section  =  ^FT 
LZ 

I,  for  circular  section  =  -^ — 


282 


PRACTICAL  STRUCTURAL  DESIGN 


I 


T 


NOTArtdH 

S'OuhiJe  l/if  merer  of  Stxt  Snelt 
H  •  Heiofit  of  Stock  exposed  to  *md. 
F-Tc&t»md  force  Zct^g 
w  ••  Hori2ontal  Kind  per  square  foot. 
8  •  Transverse  A  lonoifva'inf/ datancf 


VlH 


X 


A  concrete  chimney  is  frequently  more  economical  than  a 
brick  chimney  because  the  maximum  compressive  stress  is  greater 
than  for  brick  and  the  tension  can  be  equal  to  nfc>  in  which 
n  =  ratio  of  deformation  between  steel  and  concrete, 
fe  =  maximum  unit  compressive  stress  in  the  concrete. 
The  wind  may  come  from  any  direction    therefore,  the  rein- 
forcement must  be  equally  spaced  around  the  circumference  of 

the  shaft.  The  shaft 
may  then  be  designed 
by  trial.  Assume  a 
certain  steel  area  and 
assume  it  be  in  the 

H-Uomentfarm^^ing4^a.  fOFm     Of    a   thill    rfttg    Of 

"r-^£Sn££$7£VsB  steel-  Find  the  mo- 
ment of  inertia  and 
ascertain  how  much 
direct  load  and  bend- 
ing load  it  will  carry 
(see  page  67).  The 
steel  may  be  assumed 
to  have  a  value  of 
12,000  Ibs.  per  sq.  in. 
A  concrete  shell  is  de- 
signed to  carry  the 
direct  load  and  wind 
moment  the  steel  can- 
not carry,  and  when 
the  combined  concrete 
section  and  steel  sec- 
tion is  found,  which 
will  carry  the  direct 


Fig.  181  —  Formulae  for  Self -supporting  Steel 
Stacks 


load  and  wind  moment,  the  steel  is  placed  in  the  middle  of  the 
thickness  of  the  concrete  ring  in  the  form  of  rods  or  bars. 

Each  section  is  designed  in  this  manner  and  when  the  base  is 
reached  the  vertical  bars  are  run  into  it  a  sufficient  length  for 
anchorage. 

Self-supporting  steel  stacks  are  often  -carried  on  columns  and 
the  proper  formulas  to  use  for  this  condition  are  shown  in  Fig. 
181.  These  stacks  may  be  anchored  to  concrete  foundations 
by  means  of  bolts,  or  they  may  be  on  girders  and  anchored  by  a 


COLUMNS   AND   STRUCTURES  283 

number  of  bolts  to  rings  riveted  to  the  girders.  To  determine 
the  size  of  the  bolts  the  chimney  is  assumed  as  tending  to  over- 
turn about  one  edge  of  the  bolt  circle.  The  principle  is  that  used 
in  the  case  of  bolts  fastening  column  bases  to  footings. 

Brick  stacks  may  usually  start  with  a  thickness  of  nine  inches 
for  the  top  twenty-five  feet  and  increase  half  a  brick  in  each 
twent}r-five  feet  down.  This  is  merely  a  rule  by  which  to  deter- 
mine trial  thicknesses.  A  general  rule  for  the  top  thickness  is 
as  follows 

t  =  3  +0.4  +  0.005  H, 

in  which  t  =  thickness    in   inches    of   upper   course    (neglecting 
ornamentation). 

d  =  clear  inside  diameter  at  top  in  feet, 
H  =  height  of  stack  in  feet. 

The  thickness  of  metal  in  steel  stacks  is  governed  by  dura- 
bility as  well  as  by  strength.  The  stack  is  a  hollow  circular  can- 
tilever beam  (see  page  67)  in  which  the  weight  of  the  metal  is  of 
relatively  small  importance,  the  wind  being  the  largest  force. 
It  is  usual  to  start  with  plates  f  in.  thick  at  the  top  of  the  stack 
if  not  lined  and  some  designers  use  |  in.  plates  at  the  top  for 
lined  stacks.  Some  designers  increase  the  thickness  by  TV  in. 
each  30  or  40  feet,  while  others  increase  by  ^  in.  At  each  30  or 
40  feet  the  section  is  investigated  and  the  thickness  of  the  plate 
fixed  by  the  fiber  stress. 

To  insure  tight  joints  the  rivet  spacing  is  not  less  than  2.5 
times  the  rivet  diameter,  or  more  than  16  tunes  the  thickness  of 
the  plate.  Usually  the  rivet  spacing  is  investigated  and  deter- 
mined only  for  the  lowest  tier  of  plates  of  any  thickness.  The 
rivets  are  in  shear  due  to  bending  moment  as  well  as  ordinary 
shear. 

Not  enough  data  is  available  for  reinforced  concrete  chimneys 
to  fix  trial  thicknesses,  as  for  brick  and  steel  stacks.  The  least 
thickness  should  be  six  inches.  The  average  increase  in  thick- 
ness is  approximately  at  the  rate  of  one  inch  in  50  ft.  for  trial 
sections. 

Tanks  and  Retaining  Walls 

Fig.  182  shows  curves  for  ascertaining  the  pressure  in  bins 
and  tanks.  They  were  computed  by  the  author  when  he  was 
Chief  Engineer  of  the  Fireproof  Construction  Bureau,  Portland 


284 


PRACTICAL   STRUCTURAL   DESIGN 


Fig.  182  —  Curves  for  Designing  Tanks 


COLUMNS  AND   STRUCTURES  285 

Cement   Association,  and   have   been   printed   in   booklets   and 
several  periodicals. 

These  curves  were  computed  by  the  Janssen  formula  and 
checked  with  other  curves  and  tables.  The  pressures  are  for 
walls  of  concrete.  For  walls  of  steel  multiply  by  1.20  and  for 
walls  of  wood  multiply  by  0.95. 

When  the  depth  of  a  tank  is  less  than  the  diameter  the  surface 
of  the  slope  of  repose  of  the  material  will  pass  through  the  top 
without  intersecting  the  opposite  wall.  The  pressure  in  such 
a  case  is  similar  to  that  exerted  by  a  fluid  and  the  expression  for 
pressure  at  any  depth  is 

P  =  md, 
in  which  m  =  a  constant, 

d  =  depth  hi  feet, 

P  =  pressure  in  pounds  per  square  foot. 

The  constant  m  for  water  is  31.25,  no  matter  how  deep  the  tank 
or  what  its  diameter.  The  constants  for  common  materials  are 
shown  in  Fig.  182.  They  represent  one-half  the  weight  of  an 
equivalent  fluid,  that  is  a  fluid  which  at  any  depth  exerts  the 
same  pressure  as  the  material  considered. 

When  granular  materials  are  confined  in  deep  bins  the  opposite 
sides  come  into  play  as  soon  as  the  depth  exceeds  the  diameter. 
The  friction,  of  the  material  against  the  walls  causes  the  walls  to 
carry  some  of  the  load,  whereas  with  a  fluid  the  pressure  is  always 
normal  to  the  surface  pressed.  For  deep  bins  the  pressure  at 
any  depth  in  a  bin  or  tank  may  be  read  directly  from  the  curves. 

WTien  the  depth  of  the  bin  is  less  than  the  diameter  the  total 
pressure  against  a  vertical  strip  one  foot  wide  is 

H  =  Mffi, 
in  which  H  =  total  horizontal  pressure. 

The  overturning  moment  M  =  H  x  5- 

o 

The  foregoing  formulas  are  used  to  design  square  and  rect- 
angular tanks  and  bins,  and  retaining  walls,  holding  water  or 
granular  dry  materials. 

In  a  circular  tank  the  horizontal  pressure  is  coverted  into 
tension  in  the  circumference. 

T  =  W  x  f , 
in  which  T  =  circumferential  tension  in  a  strip  one  foot  deep, 


286  PRACTICAL  STRUCTURAL   DESIGN 

D  =  diameter  in  feet, 

W  =  the  weight  of  one  cubic  foot  of  a  fluid. 

It  is  customary  to  take  one-half  the  weight  of  the  fluid,  or 
equivalent  fluid,  and  multiply  it  by  the  diameter.  This  is  the 
constant  m  for  each  material  shown  in  Fig.  182. 

The  circumferential  tension  divided  by  the  allowable  fiber 
stress  in  the  material  gives  the  number  of  square  inches  required. 
If  the  tank  is  to  be  of  steel  the  thickness  of  the  plate  is  found  by 
dividing  the  area  by  12,  the  depth  of  the  strip.  The  proper  allow- 
ance must  then  be  made  for  rivet  holes. 

If  the  tank  is  to  be  of  reinforced  concrete  the  steel  area  must 
be  sufficient  to  carry  all  the  tension.  Sometimes  cracks  will 
open  in  concrete  walls,  and  if  the  concrete  is  relied  on  to  carry 
part  of  the  stress,  the  tensional  strength  of  the  concrete  is  lost 
with  the  first  crack  and  the  steel  immediately  carries  this  addi- 
tional stress. 

When  there  are  no  cracks  in  the  concrete  it  does  carry  part  of 
the  load,  so  the  thickness  of  the  shell  is  fixed  by  assuming  that 
the  strength  of  the  concrete  in  tension  is  150  Ibs.  per  sq.  in.  Di- 
viding the  total  stress  by  150  the  concrete  area  is  found  and 
dividing  this  by  12  the  thickness  of  the  shell  is  fixed.  It  should 
never  be  less  than  four  inches  when  first  class  experienced  work- 
men are  employed  and  six  inches  is  a  safe  enough  minimum  to 
use  for  all  tanks. 

The  shell  of  the  tank  as  thus  designed  will  carry  double  the 
tension,  part  of  which  is  carried  by  the  steel  and  part  by  the 
concrete. 

Let  A  =  total  area  =  Ac  +  As  =  Ac  +  nAc, 
in  which  Ac  =  area  of  concrete  in  square  inches, 
A8=  area  of  steel  in  square  inches, 
n  =  ratio  of  deformation  between  steel  and  concrete. 

The  thickness  of  the  concrete  multiplied  by  12  is  the  total  area 
from  which  must  be  subtracted  the  area  of  the  steel,  leaving  ACJ 
the  net  concrete  area. '  The  area  of  the  steel  is  multiplied  by  n 
and  added  to  the  net  concrete  area,  this  giving  the  area,  A,  in  the 
formula.  Dividing  the  total  stress  by  A  the  average  unit  stress 
in  tension  is  obtained  and  this  is  the  stress  on  the  concrete.  Mul- 
tiplied by  n  it  gives  the  actual  unit  tensile  stress  in  the  steel.  This 
stress  is  very  low  but  the  instant  a  crack  appears  in  the  concrete, 
thus  reducing  the  section,  the  steel  stress  is  increased  'by  an 


COLUMNS   AND   STRUCTURES  287 

amount  equal  to  n  times  the  area  of  concrete  in  the  face  of  the 
crack,  times  the  average  unit  stress.  If  the  crack  extends  through 
the  wall  the  steel  carries  all  the  tension. 

Ah1  granular  materials  have  an  angle  of  repose,  which  is  as 
follows:  Sand,  25°  to  30°,  gravel  and  broken  stone,  30°  to  40°; 
ashes,  25°  to  30°;  coal,  30°  to  45°;  grain  28°.  The  angle  of 
repose  is  generally  designated  in  formulas  by  <£. 

The  factor  k  fixes  the  ratio  of  lateral  to  vertical  pressure  and 
according  to  the  theory  of  the  ellipse  of  stress, 
,  _  1  —  sin  <£ 
~  1  +  sin  <f> 

It  has  been  determined  experimentally  for  several  materials 
and  has  a  value  of  0.6  for  grain. 

The  weight  carried  on  the  bottom  of  any  bin  is  the  t^tal  weight 
of  the  material,  minus  the  weight  carried  by  the  walls.  The 
foundations  are  designed  under  the  walls  by  taking  the  total 
weight  transferred  to  the  walls,  plus  the  weight  of  the  walls,  plus 
wind  force.  If  columns  and  girders  are  used  under  the  floor  of 
the  bin,  part  of  the  weight  on  the  bottom  is  carried  to  the  walls 
as  reaction. 

Materials  have  an  angle  of  repose  which  is  an  angle  of  slope 
assumed  by  the  surface  of  the  material  when  piled.  There  is 
also  an  angle  of  friction,  the  tangent  of  which  is  known  as  the 
coefficient  of  friction.  The  coefficient  of  friction  of  the  grains 
of  material  on  each  other  is  the  factor  k.  The  coefficient  of  fric- 
tion of  the  material  against  a  surface  confining  it  is  determined 
experimentally,  and  the  factor  C  hi  Fig.  182  is  the  coefficient  of 
friction  of  the  various  materials  against  concrete.  In  determin- 
ing the  weight  carried  by  the  walls  it  is  necessary  to  consider 
k  and  C. 

The  weight  carried  by  the  wall  on  a  strip  one  foot  wide  for 
any  depth  is,  approximately, 


in  which  w  =  weight  of  material  per  cubic  foot, 

R  =  hydraulic  radius  =  —     —  -  —  in  feet, 

h  =  depth  of  bin  in  feet, 

k  =  ratio  of  lateral  to  vertical  pressure, 

C  =  coefficient  of  friction. 


288  PRACTICAL   STRUCTURAL   DESIGN 

The  above  expression  is  only  approximate,  as  the  entire  expres- 
sion is  very  complex.  The  result,  however,  is  correct  within 
such  a  small  per  cent  that  it  is  safe  to  use  it.  The  load  on  a  verti- 
cal strip  one  foot  wide  multiplied  by  the  circumference  in  feet 
gives  the  total  vertical  weight  carried  by  the  walls,  the  remainder 
being  carried  by  the  bottom.  The  weight  on  the  bottom  is  not 
uniform,  being  in  the  form  of  an  ellipsoid,  the  bending  moment 

2      WD2 
for  which  will  be  M  =  5  H — =—  provided  the  attachment  of  the 

bottom  to  the  sides  is  good. 

The  curves  may  be  used  for  round  or  square  tanks.  In  a  round 
tank  the  pressure  is  that  on  a  square  fodt  on  the  circumference. 
For  square  tanks  it  is  the  pressure  per  square  foot  of  perimeter. 
It  may  be  used  for  rectangular  tanks  in  which  the  length  is  not 
more  than  1.5  tunes  the  breadth  by  dividing  4  times  the  area  of 
the  rectangular  tank  by  the  perimeter.  This  gives  the  diameter 
of  an  equivalent  circular  tank,  or  the  side  of  an  equivalent  square 
tank,  by  means  of  which  from  Fig.  182  can  be  obtained  the  pres- 
sure to  use  with  the  dimensions  of  the  rectangular  tank. 

Hoppered  bottoms  are  used  for  bins  as  a  rule  but  are  somewhat 
expensive  when  made  of  concrete,  on  account  of  the  formwork. 
A  common  practice  for  bins  having  tunnels  underground  is  to 
make  a  flat  bottom  and  pile  cinders,  or  damp  sand,  on  it  with  the 
surface  sloping  towards  the  discharge  hole.  The  surface  is  then 
covered  with  concrete  several  inches  thick,  generally  reinforced. 

The  pressure  against  a  retaining  wall,  and  the  overturning 
moment,  may  be  obtained  by  formula,  using  the  constants  for 
equivalent  fluid  pressure.  That  method,  however,  is  good  only 
for  a  wall  retaining  a  fill  level  with  the  top  of  the  wall.  It  is  not 
applicable  to  a  surcharged  wall,  that  is,  one  holding  a  fill  which 
extends  above  the  top  of  and  slopes  to  the  wall.  The  graphical 
method  shown  in  Fig.  182  is  a  development  of  the  Coulomb 
theory  of  a  "maximum  wedge."  According  to  this  theory  the 
fill  will  not  slip  forward  until  the  surface  is  steeper  than  the 
natural  angle  of  repose.  When  it  starts  to  slip  it  breaks  on  a 
line  approximately  halfway  between  the  angle  of  repose  and  the 
vertical,  the  wedge  ahead  of  this  line  alone  exerting  an  overturn- 
ing pressure  on  the  wall. 

In  Fig.  183  the  line  AE  represents  the  surface  slope  at  the 
angle  of  repose  <£.  The  line  El  is  the  surface  of  the  fill,  the  angle 


COLUMNS   AND   STRUCTURES 


289 


x  being  the  angle  of  surcharge.  The  length  of  the  line  AE  is 
fixed  by  the  intersection  of  the  angle  of  repose  drawn  from  the 
bottom  of  the  wall  and  the  angle  of  surcharge  drawn  from  the 
top  of  the  wall.  From  the  middle  point  of  the  line  AE  a  semicircle 
ABDE  is  drawn. 

The  angle  C  is  the  angle  of  friction  of  the  filling  against  the 
back  of  the  wall.  The  line  IH  is  drawn  at  an  angle  with  the 
back  of  the  wall  equal  to  the  sum  of  the  angle  of  repose  and 
the  angle  of  friction  C  to  an  intersection  with  the  line  AE. 

From  A  as  a  center,  with  a  radius  =  AB,  describe  an  arc  cut- 
ting AE  at  F.  Draw  FJ  parallel  with  the  line  IH.  With  radius 


Fig.  183  —  Graphical  Method  for  obtaining  Pressure  against  Retaining  Wall 

FJ  describe  an  arc  intersecting  the  line  AE  at  G  and  draw  the 
triangle  FJG. 

The  line  AJ  is  the  cleavage  line  of  the  material  and  the  area 
of  the  triangle  FJG  multiplied  by  the  weight  of  a  cubic  foot  of 
the  material  will  give  the  pressure  against  a  strip  of  the  wall 
one  foot  long.  This  pressure  is  considered  to  be  concentrated  at 
a  point  above  the  base  of  the  wall  equal  to  one-third  the  height, 
the  height  being  measured  from  the  bottom  of  the  wall,  and  not 
from  the  surface  of  the  ground.  The  direction  of  this  thrust  T 
is  not  determined,  authorities  not  agreeing.  Some  assert  that 
it  is  parallel  with  the  slope  of  the  surcharge  and  some  that  it 
makes  an  angle,  C,  with  the  back,  while  others  count  it  as  a  hori- 
zontal thrust.  In  the  figure  it  is  drawn  parallel  with  the  angle 
of  friction.  A  larger  moment  is  obtained  by  considering  the 
thrust  as  horizontal  and  a  wall  designed  to  resist  this  horizontal 


290  PRACTICAL   STRUCTURAL   DESIGN 

thrust  has  a  larger  factor  of  safety  than  one  designed  to  resist 
a  thrust  at  an  angle.  The  author  designs  retaining  walls  on  the 
assumption  of  a  horizontal  thrust. 

The  resultant  of  the  weight  of  the  wall  and  the  thrust  must 
pass  through  the  base  at  such  a  point  as  will  keep  the  toe  pres- 
sure within  the  allowable  soil  pressure  per  square  foot.  It  is 
considered  best  to  keep  the  resultant  within  the  middle  third 
of  the  wall.  In  the  figure  the  thrust  is  shown  as  meeting  a  ver- 
tical line  through  the  center  of  gravity  at  a  point  one-third  the 
height  above  the  base.  This  makes  the  thrust  line  strike  the  wall 
a  trifle  above  the  one-third  point.  With  a  horizontal  thrust  the 
application  is  exactly  one-third  above  the  base. 

The  diagram  here  given  is  independent  of  the  shape  of  the 
wall.  In  fact  a  single  line  representing  the  back  of  the  wall  could 
have  been  drawn  just  as  well.  The  diagram  merely  gives  the 
amount  of  thrust,  its  direction  and  the  point  of  application.  A 
separate  diagram  may  be  drawn  from  the  wall  if  the  work  is  to 
be  graphical,  and  only  the  thrust  line  from  this  figure  will  be 
required. 

For  a  well-built  concrete  wall  not  reinforced  the  width  of  the 
base  can  be  one-third  the  height  for  ordinary  earth.  For  a  brick 
or  well-built  cut  stone  wall  with  cement  mortar  joints  the  bottom 
width  can  be  one-third  the  height  +  1.  For  an  ordinary  stone 
or  brick  wall  the  thickness  of  the  base  should  be  at  least  one-half 
the  height.  Such  empirical  rules  make  it  very  easy  to  draw  plans 
for  walls.  With  reinforced  concrete  walls  it  is  necessary  to  know 
the  pressure  and  overturning  moment  so  the  wall  may  be  designed 
to  resist  definite  forces. 

A  reinforced  concrete  retaining  wall  is  built  in  the  shape  of  a 
capital  L.  The  weight  of  the  earth  on  the  outstanding  rear  leg  is 
counted  as  part  of  the  weight  of  the  wall,  the  back  of  the  wall 
being  vertical  and  extending  upward  from  the  rear  end  of  the 

slab.    The  coefficient  of  friction  is  k  =  ^ = — r>  and  the  angle 

1  +sm  <f> 

corresponding  to  this  is  used  in  the  graphical  analysis  instead  of 
the  angle  C. 

The  thrust  acting  at  .one-third  the  height  tends  to  overturn 
the  wall.  The  vertical  front  face  of  the  wall  must,  therefore,  be 
designed  as  a  cantilever  beam  to  resist  this  moment,  anchoring 
into  the  base.  The  small  toe  in  front  is  extended  to  widen  the 


COLUMNS   AND   STRUCTURES  291 

base  and  bring  the  toe  pressure  within  the  safe  maximum  pres- 
sure. This  projecting  toe  is  designed  as  a  cantilever  beam  resist- 
ing the  upward  pressure.  The  slab  in  the  rear  is  designed  as  a 
cantilever  beam  to  carry  the  weight  of  the  earth  on  it  which 
forms  a  part  of  the  wall. 

Sometimes  the  wall  has  counterforts  along  the  back  at  regular 
intervals,  acting  as  ties.  The  spacing  of  these  counterforts  varies 
from  an  interval  equal  to  the  height  of  the  wall  for  walls  under 
fifteen  feet  in  height  to  one-third  the  height  for  walls  thirty  feet 
in  height,  and  proportionately  for  walls  more  than  fifteen  and 
less  than  thirty  feet  high.  These  counterforts  have  rods  running 
in  them  from  the  top  of  the  wall  to  the  back  edge  of  the  bottom 
slab,  to  reinforce  them  as  cantilever  girders  carrying  the  front 
slab.  The  front  slab  is  designed  as  a  slab  with  a  span  equal  to 
the  distance  center  to  center  of  counterforts, 
and  is  reinforced  horizontally. 

The  wall  designed  as  a  cantilever  has  vertical 
reinforcement  in  the  front  wall  and  the  rein- 
forcement in  the  bottom  slab  and  toe  is  normal 
to  the  face  of  the  wall.  The  counterforted 
wall  has  vertical  and  sloping  steel  in  the  coun- 
terforts, but  the  slab  reinforcement  is  all  para- 
llel  with  the  length  of  the  wall.  Restrained  Wall 

In  Fig.  184  is  shown,  diagrammatically,  a 
retaining  wall  tied  at  the  top  and  bottom.  This  may  occur  in  the 
case  of  an  area  wall  pressing  against  a  sidewalk  at  the  top  and 
against  a  heavy  floor  at  the  bottom.  It  may  occur  as  a  wall 
pressing  against  a  foundation,  or  floor  at  the  bottom,  and  having 
a  long  girder,  or  waling,  along  the  top  held  by  ties  to  deadmen. 

The  maximum  bending  moment  is  at  a  point  =  0.58  h.     It  is 

then  M  =  0.64p/i3, 

in  which  p  =  pressure  in  pounds  per  square  foot  of  a  fluid.  For 
grain,  p  =  42;  for  stone,  p  =  48;  for  bituminous  coal,  p  =  24; 
for  anthracite  coal,  p  =  36;  for  ashes,  p  =  16;  for  sand,  p  =  48; 
for  earth,  p  =  32  (average). 

The  maximum  bending  moment  caused  'by  water  is 
M  =  4/i3. 

The  fluid  weights  are  not  actual  weights  but  merely  represent 
the  weight  that  must  be  possessed  by  a  fluid  which  would  exert 
the  same  pressure  against  a  wall  as  the  material  to  which  it  cor- 


292 


PRACTICAL   STRUCTURAL   DESIGN 


responds.  The  formulas  for  fluid  pressure  are  more  simple  than 
those  in  which  a  number  of  factors  must  be  used,  so  the  material 
is  assumed  to  act  as  a  fluid  having  a  weight  per  cubic  foot  very 
much  less  than  the  actual  weight  of  the  material. 

In  Fig.  185  three  problems  in  the  design  of  lintels  over  open- 
ings are  shown.  At  A  the  lower  opening  is  spanned  by  a  lintel 
which  carries  a  load  indicated  by  the  shaded  triangle,  which  is 
equilateral.  The  reason  this  triangular  load  is  carried  is  that 

the  brick  work  bond 
will  have  strength 
enough  to  assume  a 
form  resembling  an 
arch.  The  bending 
moment  due  to  a  tri- 
angular load  is 


Fig.  185  —  Lintels  over  Openings 


The  point  of  the 
arch  over  the  lower 
opening  is  below  the  bottom  of  the  upper  opening  a  depth  equal 
to,  or  greater  than,  one-fourth  the  span  of  the  upper  opening. 
In  the  upper  opening  there  is  a  lintel  to  carry  the  coping  wall. 
A  sixty  degree  line  drawn  from  each  upper  corner  will  intersect 
the  coping  wah1;  therefore  the  lintel  must  be  figured  to  carry  all 
the  load  above  it,  within  the  shaded  area. 

At  B  is  shown  another  case.  The  sides  of  an  equilateral  tri- 
angle will  intersect  the  bottom  of  the  upper  opening  so  it  is  com- 
mon to  assume  the  sloping  lines  at  the  side  to  connect  the  corners 
of  the  openings  as  shown.  The  triangle  over  the  top  of  the  upper 
opening  is  more  than  one-fourth  the  span  below  the  top  of  the 
wall,  so  as  some  arch  action  can  take  place  the  lintel  is  assumed 
to  carry  only  the  triangular  portion  of  the  wall.  The  author 
would  not  so  design  the  two  lintels.  The  lintel  over  the  bottom 
opening  would  be  designed  to  carry  ah1  the  load  between  the 
dotted  vertical  lines. 

At  C  a  similar  condition  is  found.  The  lintel  over  the  lowest 
opening  should  be  designed  to  carry  all  the  load  between  two 
vertical  lines  extended  from  the  upper  corners  to  the  top  of  the 
wall.  It  is  not  safe  to  assume  the  sloping  form  of  the  broken 
wall  opening  unless  there  is  a  bridging  across  sufficient  to  form 


COLUMNS  AND   STRUCTURES  293 

an  arch.  That  is  why  the  arch  is  not  assumed  to  act  unless  the 
wall  across  the  break  has  depth  enough  to  insure  it  acting  as  a 
beam  to  span  the  opening. 

In  the  three  cases,  therefore,  the  lower  lintels  must  be  designed 
to  carry  the  area  of  wall  shown  between  the  dotted  lines,  and  the 
lintels  for  the  upper  openings  are  designed  to  carry  the  small 
shaded  triangular  areas  at  B  and  C  and  the  rectangular  area 
shown  at  the  top  at  A. 

A  few  things  pertaining  to  the  design  of  buildings  have  not  been 
discussed  because  they  are  to  be  found  in  the  steel  and  lumber 
handbooks,  without  which  no  designer  can  work.  With  the 
assistance  of  those  books  the  student  should  have  no  difficulty 
in  handling  all  the  ordinary  problems  arising  in  the  design  of 
structures.  It  is  hoped  that  the  application  of  the  MOMENT 
has  been  treated  so  consistently  throughout  this  book  that  the 
student  will  have  no  difficulty  in  analyzing  any  problem  that 
may  come  up  in  his  work. 


INDEX 


Accuracy  in  drawing,  237 
Aids  to  computation,  81 
Anchorage  of  cantilever  beam,  60 
Angular  displacement  of  joint,  197 
Axis,  Neutral,  in  reinforced  concrete, 

72 
Axis,  Position  of  neutral,  63 

B 

Barlow's  tables,  62 

Bars  and  rods,  table,  148 

Bases,  Built-up,  271 

Bases,  Cast  steel,  271 

Beam,  cantilever,  Load  on,  10 

Beam  carrying  several  loads,  ends 
overhanging,  40 

Beam  carrying  several  loads,  one 
end  overhanging,  40 

Beam,  definition,  23 

Beam,  Design  of  cast  iron,  64 

Beam  on  two  supports,  Bending 
moment  at  any  point,  21 

Beam  on  two  supports,  Partially 
distributed  load,  24 

Beam  on  two  supports,  Two  equi- 
distant concentrated  loads,  25 

Beam  oa  two  supports,  Uniform  and 
several  concentrated  loads,  20 

Beam,  Several  loads  on,  15 

Beam  truss,  Hammer,  226 

Beams  and  frames,  56 

Beams  and  girders,  Deflection  in, 
79 

Beams,  Belly-rod,  119 

Beams,  Breadth  and  depth  of,  90 

Beams,  Buckling  of,  101 

Beams,  Continuous,  49,  237 

Beams,  continuous,  Maximum  mo- 
ment for,  50 


Beams,  Continuous,  with  uniform 
moment  of  inertia,  13 

Beams,  Deflection  in,  98 

Beams,  Lateral  stiffness  of,  101 

Beams,  Loads  on  overhanging,  37 

Beams  on  a  slope,  100 

Beams  on  more  than  one  support, 
Computing  reactions  on,  14 

Beams  on  two  supports,  Several  con- 
centrated loads,  20 

Beams  resting  on  two  supports,  17 

Beams,  Restrained,  46 

Beams,  Riveting  ends  of,  47 

Beams,  Shear  diagrams  for,  107 

Beams,  Single  strut,  119 

Beams,  Staying  of,  101 

Beams,  Strength  and  stiffness  of,  45 

Beams,  Tredgold's,  Thos.,  rule  for, 
101 

Beams,  Twelvetree's,  W.  N.,  rule  for, 
90 

Beams  with  uniform  stress,  118 

Bearing  stress  in  wood,  139 

Bearing  stress  in  wood,  Dewell's  for- 
mula, 140 

Bearing  stress  in  wood,  Howe's  for- 
mula, 141 

Bearing  stress  in  wood,  Jacoby's 
formula,  141 

Belgian  truss,  214 

Belly-rod  beams,  119 

Bending  moment  at  any  section  of 
beam,  Rule  for,  23 

Bending  moment,  definition,  11 

Bending  moment  for  several  loads  on 
beam,  33 

Bending  moment  for  uniformly  dis- 
tributed load  on  beam,  34 

Bending  moment  on  beam,  Graphi- 
cal solution  of,  31 


295 


296 


INDEX 


Bending  moment  on  beam  on  two 

supports,  21 
Bending  moment  on  cantilever  beam,  • 

12 
Bending  moment,  Rule  for  position 

of  maximum,  36 
Bias  and  tanks,  Curves  for  designing, 

283 

Bins    and   tanks,    Square    and   rec- 
tangular, 285 

Bins,  Hoppered  bottoms  in,  288 
Bins  or  tanks,  Friction  of  material 

on  walls  of,  286,  287 
Bins,  Overturning  moment  in,  288 
Bolts  and  fastenings  in  carpentry,  142 
Bolts,  Resistance  of,  139 
Books  for  study,  205 
Bottom  chord  steel,  Wooden  truss, 

162 

Bowstring  girders,  Stresses  in,  131 
Bowstring  trusses,  222 
Braces  in  roof  trusses,  Stresses  in, 

132 
Bracing,  Diagonal,  in  tall  buildings, 

257 

Bracket  frame,  construction,  59 
Brackets  and  bases  for  columns,  269 
Brackets  and  gusset  plates,  259 
Brick  and  stone  laid  in  cement  mor- 
tar, 264 
Brick  and  stone  walls,  Bottom  width, 

290 

Brick  stacks,  Rule  for  thickness,  283 
Buckling  of  beams,  101 
Building  frames,  Moments  and  shears 

in,  260 

Building  ordinances  in  America,  92 
Buildings,  Dead  weight  of,  261 
Buildings,  Loads  on  columns  in,  261 
Built-up  lower  chord  design,  159 
Built-up  posts,  Dewell's  experiments, 

242 
Built-up  steel  bases,  271 


Cambered  tie  rod,  Truss  with,  214 
Canopy,  Design  of  sidewalk,  59 
Cantilever  beam,  Anchorage  of,  6C 


Cantilever  beam,   Bending  moment 

on,  12 

Cantilever  beam,  Load  on,  10 
Cantilever  beam,  Moment  for  con- 
centrated load  on,  12 
Cantilever  beam,  Resisting  moment 

on,  12 

Cantilever  footings,  277 
Cantilever  trusses,  235 
Caps  for  wooden  columns,  269 
Carpentry,  Bolts  and  fastenings  in, 

142 

Carrying  capacity  of  pins,  187 
Cast  iron  beam  design,  64 
Cast  iron  lintel  design,  64 
Cast  iron  or  steel  ribbed  column  base, 

270 

Cast  iron  shoe,  175 
Cast  iron  strength,  64 
Cast  steel  bases,  271 
Cement    mortar,    Stone    and    brick 

laid  in,  264 
Center  of  gravity,  Loads  act  through, 

12 

Center  of  gravity,  Loads  at,  12 
Center  of  gravity  of  group  of  rivets, 

200 
Characteristic        points,        Fidler's 

method,  238 

Chimneys,  Design  of,  279 
Chimneys,  Stresses  in,  280 
Chimneys,  Thickness  of  reinforced 

concrete,  283 

Column  and  wall  footings,  266 
Column  bars,  Attaching  to  footings, 

275 
Column    base,    Cast    iron    or    steel 

ribbed,  270 

Column  base,  Eccentric  loads  on,  273 
Column  brackets  and  bases,  269 
Column,     Horizontal     force     acting 

on,  275 
Columns  and  girders,  Contraflexure 

point  in,  258 
Columns  and  girders,   Moments  in, 

259 
Columns  and  frames,  Wind  bracing 

on,  256 


INDEX 


297 


Columns  and  posts,  Distinction  be- 
tween, 241 

Columns  and  structures,  241 
Columns,   Area  of  floor  served   by 

interior,  262 

Columns,  Caps  for  wooden,  269 
Columns,  Eccentric  loads  on,  253 
Columns,  Euler's  formula,  243 
Columns,  Foundations  under,  276 
Columns,  Gordon's  formula,  243 
Columns  in  buildings,  Loads  on,  261 
Columns,  Methods  of  fixing,  252 
Columns,  Rankine's  formula,  243 
Columns,  Wrought  iron,  251 
Compound  beams,  110 
Compression  pieces,  Built-up,  163 
Computation,  Aids  to,  81 
Computing  reactions,  18 
Concentrated  loads,  232 
Concentrated    loads    on    beams    on 

two  supports,  20 

Concrete  beam,    Moment  of   resist- 
ance, 74 

Connections  and  joists,  136 
Construction  of  parabola,  28 
Continuous  beams,  237 
Continuous   beams,    Maximum   mo- 
ment for,  50 
Contraflexure  point  in  columns  and 

girders,  258 

Contraflexure,  Point  of,  41,  46 
Counterforts  in  retaining  walls,  291 
Couple,  definition,  63 
Cover  plates  and  stiffeners  for  plate 

girders,  113 

Cover  plates,  definition,  114 
Cover  plates,  Proportioning,  117 
Cracks  in  timber  framework,  144 
Crane  loads  on  girders,  36 
Crescent  truss,  223 
Curves  for  designing  bins  and  tanks, 
283 


Dead  load,  definition,  10 
Dead  weight  of  buildings,  261 
Deck  truss,  Pratt,  222 
Deck  truss,  Warren,  220 


Deflection  formulas,  80 

Deflection  in  beams  and  girders,  79, 

98 

Deformation,  Ratio  of,  71 
Design  of  beams,  Problems  in,  85 
Design  of  chimneys,  279 
Design  of  floors,  88 
Design  of  riveted  joints,  192 
Design  of  rods,  146 
Design  of  tank  towers,  277 
Designers  and  salesmen,  54 
Designing  footings,  266 
Designs  for  joints  and  fastenings,  143 
Details  for  wood  truss,  142 
Dewell,  H.  D.,  Experiments  of,  138 
Diagonal  bracing  in  tall  buildings,  257 
Distributed  equivalent  loads,  43 
Distributed  loads  on  footings,  262 
Distribution  of  pressure  on  bearing 

faces,  170 

Dividing  line  into  equal  parts,  27 
Double  shear  rivets,  191 
Double  strut  belly-rod  beam,  120 
Drawing,  Accuracy  in,  237 
Duchemin's  formula  for  wind  pres- 
sure, 227 


Eccentric  loading,  Flexure  stress  due 

to,  254 

Eccentric  loads  on  column  base,  273 
Eccentric  loads  on  columns,  253 
Eccentric  loads  on  footings,  271 
Elastic  limit,  definition,  69 
Employment  of  engineers,  55 
End  joint,  Design  of,  173 
End  stiffeners,  113 
Engineers,  Employment  of,  55 
Equivalent  distributed  loads,  43 
Euler's  straight-line  formula,  246 
Eye-bars,  186 

F 

Fabrication  standards  for  steel,  195 
Factor  of  safety,  definition,  112 
Fan  trusses,  134 

Faulty  design,  Effect  of,  on  framed 
structures,  196 


298 


INDEX 


Fiber  stresses  in  wood,  91,  164 

Fiber  stress  of  yellow  pine,  104 

Fink  trusses,  134,  214 

Fink   truss   problem,   Wrigley's   so- 
lution, 217 

Fish-plate  splice,  153 

Flexure  in  pins,  formula,  186 

Flexure  stress  due  to  eccentric  load- 
ing, 254 
^Flitch  plate  girders,  111 

Floor,  Area  of,   served  by   interior 
columns,  262 

Floor,  definition,  123 

Floor,  Description  of  laminated,  98 

Floors,  Design  of,  88 

Fluid  weight,  291 

Foot  pounds,  definition,  12 

Footings,  Attaching  column  bars  to, 
275 

Footings,  Cantilever,  277 

Footings,  Designing,  266 

Footings,  Distributed  loads  on,  262 

Footings,  Eccentric  loads  on,  271 

Footings,  Reinforced,  268 

Footings,  Wall  and  column,  266 

Force,  Triangles  of,  62 

Forces,  Parallelogram  of,  207 

Forces,  Polygon  of,  208 

Forces,  Triangle  of,  208 

Formula     for    self-supporting    steel 
stacks,  282 

Formulas,  definition,  9 

Foundations  of  water  towers,  279 

Foundations  under  columns,  276 

Frame,  stress  at  any  point,  58,60 

Framed  structures,  Effect  of  faulty 
design,  196 

Frames  are  forces,  56 

Frames  with  inclined  struts,  57 

Free  designing,  53 

Friction  in  riveted  joints,  189 

Friction  of  material  on  walls  of  bins 
or  tanks,  285,  287 


Girder,  definition,  23 
Girders  and  trusses,  102 
Girders,  Crane  loads  on,  36 


Girders,  Flitch  plate,  111 
Girders,  Moving  loads  on,  34 
Girders,  Wheel  loads  on,  36 
Graphical   method   for   constructing 

parabola,  31 
Graphical  method  for  finding  wind 

stresses  and  reactions,  229 
Graphical  methods  for  moment  and 

shear  in  beams,  14 
Graphic  statics,  207 
Graphic  statics,  King  truss  forces  by, 

210 
Graphic  statics,  Queen  truss   forces 

by,  212 
Gusset  plates  and  brackets,  259 


Hammer  beam  truss,  226 
Hangers  and  stirrups,  293 
Heel,  Roof  truss,  203 
Hog  back  truss,  224 
Hoppered  bottoms  in  bins,  288 
Horizontal  force  acting  on  column, 

275 

Horizontal  shearing  stresses,  75 
Howe  truss,  Development  of,  123 


Impact,  definition,  210 

Inch  and  foot  pounds,  Moments     in, 

12 

Inch  pounds,  definition,  12 
Inclined  reactions  from  wind,  228 
Inertia,  Moment  of,  66 
Inflection,  Point  of,  41,  46 
Interior  columns,  Area  of  floor  served 

by,  262 

Intermediate  joints  in  trusses,  181 
Intermediate     joints,     Methods     of 

framing,  182 
Internal  forces  and  resisting  moment, 

56 
Internal  forces,  definition,  56 

J 

Joint,  Angular  Displacement  of,  197 
Joint  details  for  trusses  of  wood,  142 
Joints  and  fastenings,  Designs  for,  143 


INDEX 


299 


Joints,  definition,  23 
Joints  in  trusses,  Intermediate,  181 
Joints  in  wood,  Designing,  167 
Joints,  Locating,  in  trusses,  131 
Joints,  Riveted  vs.  pin   connections, 

185 

Joists  and  connections,  136 
Joists,  End  bearing  of,  296 
Joists,  Design  of  reinforced,  102 


Ketchum's      recommendations      for 

snow  load,  233 
King  truss,  132 
King  truss  forces  by  graphic  statics, 

210 

Knee  brace  connection,  204 
Knuckle  plate  connection,  204 


Lag  screws,  Resistance  of,  138 

Laminated  floor,  description,  98 

Lateral  deflection  and  eccentric  load- 
ing, 101 

Lateral  stays  for  steel  beams,  101 

Lateral  stiffness  of  beams,  101 

Lattice  truss,  128 

Length  of  spans,  13,  42 

Lettering  spaces  in  truss  diagrams, 
Mr.  Bow's  system,  130 

Lever  and  moment,  11 

Lewis  and  Morton's  wooden  high- 
way bridge  design,  162 

Line,  Dividing,  into  equal  parts,  27 

Lintel,  Cast  iron,  design,  64 

Lintels  over  openings,  three  designs, 
292 

Live  load,  definition,  10 

Loading  of  columns,  Conditions  af- 
fecting end,  252 

Load  on  beam,  Bending  moment  for 
uniformly  distributed,  34 

Load  on  beam  on  two  supports,  Par- 
tially distributed,  24 

Load  on  cantilever  beam,  Moment 
for  concentrated,  12 

Loads  act  through  center  of  gravity, 
12 


Loads  at  center  of  gravity,  12 
Loads,    Effect    of    combining    con- 
centrated and    uniformly  distrib- 
uted, 16,  25 
Loads   on   beam,    Bending   moment 

for  several,  33 
Loads   on   beam   on   two   supports, 

Two  equidistant  concentrated,  25 
Loads  on  beam,  Several,  15 
Loads  on  columns  in  buildings,  261 
Loads    on    columns,    Johnson's    for- 
mula, 255 

Loads  on  overhanging  beams,  37 
Loads  on  roofs,  135 
Loads  on  span,  Two  unequal,  37 
Lower  chord,  Built-up  design,  159 
Lower  chord,  Design  of,  162 

M 

Machine  driving,  Rivet  clearance  for, 

199 
Maximum    bending    moment,    Rule 

for  position  of,  36 
Maximum,    minimum    and   reversed 

stresses,  232 
Maximum    moment    for    continuous 

beams,  50 
Maximum  shear  on  wooden  beams, 

97 
Metal  in  steel  stacks,  Thickness  of, 

233 

Methods  of  fixing  columns,  252 
Methods    of    framing    intermediate 

joints,  182 

Mill-constructed  building,  104 
Modulus  of  elasticity,  definition,  70 
Modulus  of  rupture,  78 
Moisture  in  woodwork  joints,  144 
Moment  and  shear  in  beams,  Graphi- 
cal methods  for,  14 
Moment  and  shear,  Relation  between, 

15,42 

Moment  curve  for  plate  girders,  116 
Moment   for  concentrated   load   on 

cantilever  beam,  12 
Moment,  Importance  of  principles  of, 

11 
Moment  of  inertia,  66 


300 


INDEX 


Moment  of  inertia,  Continuous  beams 

with  uniform,  13 
Moment  of  inertia,  Method  of  finding, 

66 

Moment  of  resistance  defined,  62,  65 
Moment    of   resistance   of    concrete 

beam,  74 
Moments    and    shears    in    building 

frames,  260 

Moments,  definition,  10 
Moments  in  columns  and  girders,  259 
Moments  in  inch  and  foot  pounds, 

12 
Moments     on     continuous     beams, 

Church's  method,  238 
Moments  on  continuous  beams,  Du 

Bois'  method,  238 
Moments  on  continuous  beams,  Fid- 

ler's  method,  238 
Moments,  Unbalanced,  53 
Moving    loads    on    girders,  34 


Nails,  Resistance  of,  106,  138 
National     Lumber     Manufacturers' 

Association,  80 
Neutral  axis  in  reinforced  concrete, 

72 

Neutral  axis,  Position  of,  63 
Neutral  plane,  definition,  62 


Ordinate    method    for    constructing 

parabola,  30 
Overturning  moment  in  bins,  288 


Parabola,  Construction  of,  28 

Parabola,  Graphical  method  for  con- 
structing, 31 

Parabola,  Ordinate  method  for  con- 
structing, 30 

Parallelogram  of  forces,  207 

Pearson's  formula  for  wind  pressure, 
227 

Pin  connections,  185 

Pin  joints,  Design  of,  186 

Pins,  carrying  capacity,  187 


Plate  connection,  Knuckle,  204 
Plate  girder  flanges,  Spacing  of  rivets 

in,  115,  117 
Plate  girders,  Cover  plates  and  stif- 

feners  for,  113 
Plate  girders,  Design  of,  112 
Plate,  steel,  Table  of,  150 
Point  of  contraflexure,  41,  46 
Point  of  inflection,  41,  46 
Point  of  reverse  moment,  41,  46 
Polygon  of  forces,  208 
Portland  Cement  Association,  74 
Posts  and  columns,  Distinction  be- 
tween, 241 
Pratt  deck  truss,  222 
Pratt  through  truss,  221 
Pratt  truss,  Development  of,  121 
Pressure  against  retaining  walls,  289 
Problems  in  design  of  beams,  85 
Proportioning  struts  or  compression 

members,  252 
Purlins,  definition,  23 
Pythagoras,  Rule  of,  57 


Queen  truss,  132 

Queen  truss  forces  by  graphic  statics, 
212 


Radii  of  gyration,  Approximate,  245, 
250 

Reactions  and  wind  stresses,  Graphi- 
cal method  for  finding,  229 

Reactions  for  continuous  beams,  52 

Reactions  on  supports,  13 

Rectangular  reinforced  concrete 
beams,  formula,  74 

Reinforced  concrete  beams,  71 

Reinforced  concrete  beams,  Failure 
in  diagonal  shear  of,  77 

Reinforced  concrete  chimneys,  Thick- 
ness of,  283 

Reinforced  concrete  design,  Slide 
rules  for,  81 

Reinforced  concrete  footings,  268 

Reinforced  concrete,  Neutral  axis  in, 
72 


INDEX 


301 


Reinforced  concrete  retaining  wall, 
290 

Reinforced  joists,  Design  of,  102 

Reinforcing  planks,  Length  of,  106 

Relation  between  moment  and  shear, 
15,  42 

Resistance,  Development  of  mo- 
ment of,  65 

Resistance,  Moment  of,  defined,  63 

Resistance  of  lag  screws,  138 

Resistance  of  nails,  106,  138 

Resistance  of  wood  screws,  138 

Resisting  forces,  Action  of,  56 

Resisting  moment  and  internal  forces, 
56 

Resisting  moment,  definition,  1 1 

Resisting  moment  on  cantilever  beam, 
12 

Restrained  beams,  46 

Retaining  wall  tied  at  top  and  bottom, 
291 

Retaining  wall,  Reinforced  concrete, 
290 

Retaining  walls  and  tanks,  283 

Retaining  walls,  Counterforts  in,  291 

Retaining  walls,  Pressure  against,  289 

Reverse  moment,  Point  of,  41,  46 

Rivet  clearance  for  machine  driving, 
199 

Rivet  spacing,  Standards  for,  195 

Rivets,  and  riveting,  188 

Rivets,  Double  shear,  191 

Rivets,  Finding  center  of  gravity  of 
group  of,  200 

Rivets  in  metal,  106 

Rivets  in  plate  girder  flanges,  Spac- 
ing, 115,  117 

Rivets,  Shearing  and  bearing  value 
of,  table,  190 

Riveted  connections,  185 

Riveting  ends  of  beams,  47 

Riveted  joints  vs.  pin  connections, 
185 

Rivets,  Single  shear,  191 

Rivets,  Table  of,  190 

Rivets,  Strength  of,  190 

Rods,  Design  of,  146 

Rods  and  bars,  table,  148 


Roof,  curved,  Wind  on,  234 

Roof  loads,  135 

Roof  truss  heel,  203 

Roof  trusses,  Stresses  in,  132 

Roof  trusses,  Tables  of  stresses,  229 

Rule  for  bending  moment  at  any 
section  of  beam,  23 

Rule  for  locating  point  of  zero  shear, 
36 

Rule  for  position  of  maximum  bend- 
ing moment,  36 

Rule  of  Pythagoras,  57 

Rupture,  Modulus  of,  78 

S 

Safety,  Definition  of  factor  of,  12 

Scissors  truss,  225 

Screw  end  bars,  table,  147 

Secondary  stresses  in  framed  struc- 
tures, 196 

Section  modulus,  definition,  66 

Section  modulus  for  T-sections, 
Method  of  finding,  66 

Shear  and  moment,  Relation  be- 
tween, 15,  42 

Shear,  definition,  14 

Shear  diagrams  for  beams,  107 

Shear  in  wooden  beams,  78 

Shear,  zero,  Rule  for  locating  point 
of,  36 

Shearing  and  bearing  value  of  rivets, 
table,  190 

Shear-pin  splice,  155 

Shearing  resistance,  75 

Shearing  stresses,  horizontal,  75 

Shearing  stress  in  steel,  75 

Shears  and  moments  in  building 
frames,  260 

Shed  roof  truss,  224 

Shop-driven  rivets,  189 

Sidewalk  canopy  design,  59 

Signs  used  for  stresses,  135 

Single  shear  rivets,  191 

Single  strut  beams,  119 

Slide  rules,  81 

Slide  rules  for  reinforced  concrete 
design,  81 

Slope,  Beams  on,  100 


302 


INDEX 


Smoley's   tables    for   structural   de- 
signers, 62 

Snow  load,  Ketchum's  recommenda- 
tions, 233 

Span,  Length  of,  42 
Span,  Two  unequal  loads  on,  37 
Specifications  for  structural  steel,  195 
Spikes,  see  Nails,  106 
Splice,  shear-pin,  155 
Stacks,  brick,  Rule  for  thickness,  283 
Stair  and  rafter  stringers,  100 
Standards  for  rivet  spacing,  195 
Standards  for  steel  work,  195 
Statics,  Graphic,  207 
Steel  beams,  Lateral  stays  for,  101 
Steel   column   formula   used   in   the 

United  States,  248 
Steel  handbooks,  85,  87 
Steel  plate,  Table  of,  150 
Steel  salesmen  designers,  54 
Steel,  Shearing  stress  in,  75 
Steel  stacks,   Formula  for    self-sup- 
porting, 282 

Steel  work,  Standards  for,  195 
Stiffness  of  wood  beams,  101 
Stirrups  and  hangers,  93 
Stirrup  sizes  and  capacities,  table,  96 
Stone  and  brick  laid  in  cement  mor- 
tar, 264 
Stone  and  brick  walls,  Bottom  width, 

290 

Straight-line  formula,  Euler's,  246 
Straight-line  formula,  Winslow's,  242 
Strength  and  stiffness  of  beams,  45 
Stress  at  any  point  in  frame,  58,  60 
Stress  in  water  towers,  278 
Stress  strain  diagram,  69 
Stresses  in  bowstring  girders,  131 
Stresses  in  braces  in  roof  trusses,  132 
Stresses  in  chimneys,  280 
Stresses  in  roof  trusses,  132 
Stresses  in  towers,  277 
Stresses  in  trusses,  124 
Stresses  in  Warren  truss,  128 
Stresses,  Secondary,  in  framed  struc- 
tures, 196 

Stresses,     Secondary,     in     wooden 
3,  183 


Stresses,  Signs  used  for,  135 
Stringers,  Stair  and  rafter,  100 
Structural   steel,    Specifications   for, 

195 

Structure,  definition,  9 
Structures  and  columns,  241 
Struts  or  compression  members,  Pro- 
portioning, 252 

Stub  tabled  fish-plate  joint,  158 
Supports,  Beams  resting  on  two,  17 


Tabled  fish-plate  splice,  156 
Tables  of  stresses  for  roof  trusses,  229 
Tank,  iron  and  steel,  table,  149 
Tank  tower  design,  277 
Tank  walls,  Weight  against,  287 
Tanks  and  bins,  Curves  for  design- 
ing, 283 

Tanks  and  bins,  Square  and  rectan- 
gular, 285 

Tanks  and  retaining  walls,  283 
Tension  in  rivets,  189 
Three-moment  theorem,  238 
Through  truss,  Pratt,  221 
Through  truss,  Warren,  220 
Tie  and  strut,  Design  of  combined, 

153 

Timber  framework,  Cracks  in,  144 
Timbers,  Working  value  of,  139 
Top  chords  of  trusses,  Making  joints 

for,  162 

Towers,  Stresses  in,  277 
Tredgold's,  Thos.,  rule  for  beams,  101 
Triangle  of  forces,  208 
Triangle  the  strongest  frame,  56 
Triangles  of  force,  62 
Truss  a  skeleton  beam,  121,  125,  126 
Truss,  Crescent,  223 
Truss,  Hog  back,  224 
Truss,  Lattice,  128 
Truss,  Scissors,  225 
Truss,  Shed  roof,  224 
Truss  with  cambered  tie  rod,  214 
Truss    with    subvertical    and    sub- 
diagonal  ends,  127 
Trussed  beams,  118 
Trusses  and  girders,  102 


INDEX 


303 


Trusses,  Cantilever,  235 

Trusses,  Coefficient  for,  122,  123 

Trusses,  definition,  121 

Trusses,  Intermediate  joints  in,  181 

Trusses,  Locating  joints  in,  131 

Trusses  of  wood,  Details  for,  142 

Trusses,  Ratio  of  depth  to  span  in, 

130 
T-sections,  Method  of  finding  section 

modulus,  66 
Twelvetree's,  W.  N.,  rule  for  beams,  90 


Unbalanced  moments,  53 

Unequal  loading,  218 

Uniform  and  concentrated  loads  on 

beam  on  two  supports,  20 
Uniform  stress  in  beams,  118 
United  States,  Steel  column  formula 

used  in,  248 

Unit  moment  of  resistance,  74 
Unit  shear,  75 
Upper  chord,  Design  of,  162 


Vertical  shearing  stresses,  75 

W 

Wall  and  column  footings,  266 
Walls  of  bins  or  tanks,  Friction  of 

material  on,  285,  287 
Warren  deck  truss,  220 
Warren  through  truss,  220 
Warren  truss,  Stresses  in,  128 
Washers,  Design  of,  150 
Water  towers,  Foundations  of,  279 
Water  towers,  Stress  in,  278 
Weight  against  tank  walls,  287 
West   Coast   Lumbermen's  Associa- 
tion, 80 


Wheel  loads  on  girders,  36 

Wind,  Action  of,  10 

Wind  bracing  on  columns  and  frames, 

256 

Wind  force,  227 

Wind,  Inclined  reactions  from,  228 
Wind  on  a  curved  roof,  234 
Wind  pressure,  Duchemin's  formula, 

227 
Wind   pressure,    Pearson's   formula, 

227 

Wind  stresses  and  reactions,  Graphi- 
cal method  for  finding,  229 
Winslow's  straight-line  formula,  242 
Wood  beams,  Stiffness  of,  101 
Wood,  Deflection  of,  49 
Wood,  Fiber  stresses  in,  91 
Wood  screws,  Resistance  of,  138 
Wood  ties  in  framework,  138 
Wood  truss  details,  142 
Wooden  beams,  Maximum  shear  on, 

97 

Wooden  beams,  Shear  in,  78 
Wooden  beams,   Strength  in  shear, 

formula,  78 
Wooden  highway  bridge,   design  of 

H.  C.  Lewis  and  E.  R.  Morton,  162 
Wooden  trusses,  Secondary  stresses 

in,  183 

Wrigley's  Fink  truss  problem,  217 
Wrought  iron  columns,  251 


Yellow  pine,  Fiber  stress  of,  104 
Yellow  Pine  Manufacturers'  Associa- 
tion, 80 


Zero  shear,  Rule  for  locating  point 
of,  36 


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